NCERT Class 11 Maths Exercise 3.3 Solutions – Trigonometric Functions (Chapter 3)

Concept Used

This problem is based on the standard trigonometric values of special angles. Angles such as \(30^\circ\), \(45^\circ\), and \(60^\circ\) (or \(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\)) frequently appear in trigonometric identities and simplifications.

Important standard values:

\(\sin 30^\circ = \dfrac{1}{2}\)
\(\cos 60^\circ = \dfrac{1}{2}\)
\(\tan 45^\circ = 1\)

By substituting these known values into the expression, the identity can be verified through simple algebraic simplification.

Solution Roadmap

Convert radian angles into familiar degree angles.
Substitute standard trigonometric values.
Square each term carefully.
Simplify the resulting arithmetic expression.
Verify that LHS equals RHS.

Illustration of Standard Angles

Solution

We need to prove

\[ \sin ^{2}\dfrac{\pi}{6}+\cos ^{2}\dfrac{\pi}{3}-\tan ^{2}\dfrac{\pi}{4}=-\dfrac{1}{2} \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &= \sin ^2\dfrac{\pi}{6} + \cos ^2\dfrac{\pi}{3} - \tan ^2\dfrac{\pi}{4} \end{aligned} \]

Using standard trigonometric values:

\[ \begin{aligned} \sin 30^\circ &= \dfrac{1}{2}, \\ \cos 60^\circ &= \dfrac{1}{2}, \\ \tan 45^\circ &= 1 \end{aligned} \]

Substituting these values:

\[ \begin{aligned} \text{LHS} &= \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 - 1 \\ &= \dfrac{1}{4} + \dfrac{1}{4} - 1 \\ &= \dfrac{1}{2} - 1 \\ &= -\dfrac{1}{2} \end{aligned} \]

Thus,

\[ \text{LHS} = \text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Questions involving evaluation of trigonometric expressions using special angles are common in CBSE and other state board examinations.
Competitive examinations such as JEE, NEET, BITSAT and NDA frequently test quick recall of standard trigonometric values.
Mastery of these values helps students simplify complex trigonometric identities and solve problems involving angles in radians.
Such problems strengthen conceptual understanding required for advanced topics like trigonometric equations, identities, and calculus applications.

Concept Used

This problem involves evaluation of trigonometric expressions using special angles and quadrant properties of trigonometric functions.

The angles used here are

\(\dfrac{\pi}{6} = 30^\circ\)
\(\dfrac{\pi}{3} = 60^\circ\)
\(\dfrac{7\pi}{6} = 210^\circ\)

The angle \(\dfrac{7\pi}{6}\) lies in the third quadrant where sine is negative. Since cosecant is the reciprocal of sine,

\[ \sin\left(\dfrac{7\pi}{6}\right)=-\dfrac{1}{2} \]

Therefore,

\[ \text{cosec}\left(\dfrac{7\pi}{6}\right)=-2 \]

When squared, the negative sign disappears.

Solution Roadmap

Convert radian angles into familiar degree angles.
Use standard trigonometric values.
Apply quadrant rules to determine the sign of sine.
Evaluate cosecant using reciprocal identity.
Substitute values and simplify algebraically.

Quadrant Illustration for \(7\pi/6\)

Solution

We need to prove

\[ 2\sin ^{2}\dfrac{\pi}{6}+\text{cosec}^{2}\dfrac{7\pi}{6}\cdot \cos ^{2}\dfrac{\pi}{3}=\dfrac{3}{2} \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=2\sin ^{2}\dfrac{\pi}{6}+\text{cosec}^{2}\dfrac{7\pi}{6}\cdot \cos ^{2}\dfrac{\pi}{3} \end{aligned} \]

Using standard values

\[ \begin{aligned} \sin\dfrac{\pi}{6}=\dfrac{1}{2}, \\ \cos\dfrac{\pi}{3}=\dfrac{1}{2} \end{aligned} \]

Also,

\[ \sin\dfrac{7\pi}{6}=-\dfrac{1}{2} \]

Therefore,

\[ \text{cosec}\dfrac{7\pi}{6}=-2 \]

Substituting the values:

\[ \begin{aligned} \text{LHS} &=2\left(\dfrac{1}{2}\right)^2+(-2)^2\left(\dfrac{1}{2}\right)^2 \\ &=2\left(\dfrac{1}{4}\right)+4\left(\dfrac{1}{4}\right) \\ &=\dfrac{1}{2}+1 \\ &=\dfrac{3}{2} \end{aligned} \]

Thus,

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Such questions test the student's knowledge of standard trigonometric values and quadrant rules.
CBSE and other board examinations frequently include identities requiring substitution of special angles.
Competitive examinations like JEE Main, NEET, BITSAT and NDA often require quick evaluation of expressions involving angles such as \(30^\circ\), \(45^\circ\), \(60^\circ\), \(210^\circ\) etc.
Understanding quadrant signs is extremely important in solving trigonometric equations and inverse trigonometry problems in higher mathematics.

Concept Used

This problem uses the standard trigonometric values of special angles together with reciprocal identities of trigonometric functions.

Important relationships used in this question are

\(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
\(\cot\theta = \dfrac{1}{\tan\theta}\)
\(\text{cosec}\ \theta = \dfrac{1}{\sin\theta}\)

The angle \(\dfrac{\pi}{6}\) corresponds to \(30^\circ\) and \(\dfrac{5\pi}{6}\) corresponds to \(150^\circ\). The angle \(150^\circ\) lies in the second quadrant where sine is positive.

Important standard values used:

\(\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}\)
\(\cot\dfrac{\pi}{6} = \sqrt{3}\)
\(\sin\dfrac{5\pi}{6} = \dfrac{1}{2}\)
\(\text{cosec}\ \dfrac{5\pi}{6} = 2\)

Solution Roadmap

Identify the special angles present in the expression.
Recall the standard trigonometric values of \(30^\circ\) and \(150^\circ\).
Use reciprocal identities for cotangent and cosecant.
Substitute values carefully.
Simplify the algebraic expression.

Unit Circle Illustration for \(30^\circ\) and \(150^\circ\)

Solution

We need to prove

\[ \cot ^2\dfrac{\pi}{6}+\text{cosec}\dfrac{5\pi}{6}+3\tan ^2\dfrac{\pi}{6}=6 \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &= \cot ^2\dfrac{\pi}{6}+\text{cosec}\dfrac{5\pi}{6}+3\tan ^2\dfrac{\pi}{6} \end{aligned} \]

Using standard values

\[ \begin{aligned} \cot\dfrac{\pi}{6}&=\sqrt{3}, \\ \tan\dfrac{\pi}{6}&=\dfrac{1}{\sqrt{3}}, \\ \text{cosec}\ \dfrac{5\pi}{6}&=2 \end{aligned} \]

Substitute these values:

\[ \begin{aligned} \text{LHS} &=(\sqrt{3})^2+2+3\left(\dfrac{1}{\sqrt{3}}\right)^2 \\ &=3+2+3\left(\dfrac{1}{3}\right) \\ &=3+2+1 \\ &=6 \end{aligned} \]

Thus,

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Problems based on standard trigonometric values frequently appear in CBSE and state board examinations.
Competitive examinations such as JEE Main, NEET, BITSAT and NDA often require rapid evaluation of trigonometric expressions involving special angles.
Strong familiarity with angles \(30^\circ, 45^\circ, 60^\circ, 150^\circ\) helps students simplify complex identities quickly.
These concepts are foundational for later topics such as trigonometric equations, inverse trigonometry and calculus.

Concept Used

This question evaluates a trigonometric expression using standard values of special angles from the unit circle.

The angles appearing in the expression correspond to familiar degree measures:

\(\dfrac{3\pi}{4} = 135^\circ\)
\(\dfrac{\pi}{4} = 45^\circ\)
\(\dfrac{\pi}{3} = 60^\circ\)

Important trigonometric values used:

\(\sin 135^\circ = \dfrac{1}{\sqrt{2}}\)
\(\cos 45^\circ = \dfrac{1}{\sqrt{2}}\)
\(\cos 60^\circ = \dfrac{1}{2}\)
\(\sec 60^\circ = \dfrac{1}{\cos 60^\circ} = 2\)

The angle \(135^\circ\) lies in the second quadrant where sine is positive and cosine is negative. Since the sine term is squared in this expression, the result remains positive.

Solution Roadmap

Convert radian angles into familiar degree angles.
Recall standard trigonometric values from the unit circle.
Use reciprocal identity for secant.
Square the trigonometric values carefully.
Simplify the resulting arithmetic expression.

Unit Circle Illustration of \(45^\circ\), \(60^\circ\), and \(135^\circ\)

Solution

We need to prove

\[ 2\sin ^2\dfrac{3\pi}{4}+2\cos ^2\dfrac{\pi}{4}+2\sec ^2\dfrac{\pi}{3}=10 \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=2\sin ^2\dfrac{3\pi}{4}+2\cos ^2\dfrac{\pi}{4}+2\sec ^2\dfrac{\pi}{3} \end{aligned} \]

Using standard trigonometric values:

\[ \begin{aligned} \sin\dfrac{3\pi}{4}&=\dfrac{1}{\sqrt{2}}, \\ \cos\dfrac{\pi}{4}&=\dfrac{1}{\sqrt{2}}, \\ \sec\dfrac{\pi}{3}&=2 \end{aligned} \]

Substitute these values.

\[ \begin{aligned} \text{LHS} &=2\left(\dfrac{1}{\sqrt{2}}\right)^2 +2\left(\dfrac{1}{\sqrt{2}}\right)^2 +2(2)^2 \\ &=2\left(\dfrac{1}{2}\right) +2\left(\dfrac{1}{2}\right) +2\times4 \\ &=1+1+8 \\ &=10 \end{aligned} \]

Thus,

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Evaluation of trigonometric expressions using standard angles is a frequently tested concept in CBSE and other state board examinations.
Competitive examinations such as JEE Main, NEET, BITSAT and NDA often require quick simplification of expressions involving angles like \(45^\circ\), \(60^\circ\) and \(135^\circ\).
Such problems strengthen familiarity with the unit circle and reciprocal trigonometric functions like secant and cosecant.
Mastery of these values greatly helps in solving trigonometric identities, equations and calculus-based problems in higher mathematics.

Concept Used

Angles such as \(75^\circ\) and \(15^\circ\) are not among the basic standard angles. However, they can be expressed as combinations of known angles like \(30^\circ\) and \(45^\circ\).

To evaluate such angles we use compound angle identities.

Angle addition identity:

\[ \sin(A+B)=\sin A \cos B+\cos A \sin B \]

Angle subtraction identity:

\[ \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \]

These identities allow us to express unknown angles using standard trigonometric values.

Solution Roadmap

Express the required angles as sums or differences of \(30^\circ\) and \(45^\circ\).
Apply the appropriate compound angle identity.
Substitute standard trigonometric values.
Simplify the algebraic expression carefully.

Angle Combination Illustration

Solution

(i) Finding the value of \(\sin 75^\circ\)

\[ \begin{aligned} \sin 75^\circ &=\sin(45^\circ+30^\circ) \end{aligned} \]

Using the identity

\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]

\[ \begin{aligned} \sin 75^\circ &=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ \end{aligned} \]

Substitute standard values

\[ \begin{aligned} \sin45^\circ &=\dfrac{1}{\sqrt2},\\ \cos30^\circ &=\dfrac{\sqrt3}{2},\\ \sin30^\circ &=\dfrac{1}{2} \end{aligned} \]

\[ \begin{aligned} \sin75^\circ &=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2} +\dfrac{1}{\sqrt2}\cdot\dfrac12 \\ &=\dfrac{\sqrt3}{2\sqrt2}+\dfrac{1}{2\sqrt2} \\ &=\dfrac{\sqrt3+1}{2\sqrt2} \end{aligned} \]

Therefore

\[ \sin75^\circ=\dfrac{\sqrt3+1}{2\sqrt2} \]

(ii) Finding the value of \(\tan 15^\circ\)

\[ \begin{aligned} \tan15^\circ &=\tan(45^\circ-30^\circ) \end{aligned} \]

Using the identity

\[ \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \]

\[ \begin{aligned} \tan15^\circ &=\dfrac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ} \end{aligned} \]

Substituting standard values

\[ \tan45^\circ=1,\qquad \tan30^\circ=\dfrac{1}{\sqrt3} \]

\[ \begin{aligned} \tan15^\circ &=\dfrac{1-\dfrac{1}{\sqrt3}}{1+\dfrac{1}{\sqrt3}} \end{aligned} \]

Multiply numerator and denominator by \(\sqrt3-1\)

\[ \begin{aligned} \tan15^\circ &=\dfrac{\sqrt3-1}{\sqrt3+1}\times\dfrac{\sqrt3-1}{\sqrt3-1}\\ &=\dfrac{(\sqrt3-1)^2}{3-1}\\ &=\dfrac{3+1-2\sqrt3}{2}\\ &=2-\sqrt3 \end{aligned} \]

Thus

\[ \tan15^\circ=2-\sqrt3 \]

Significance for Board and Competitive Examinations

Deriving trigonometric values of non-standard angles such as \(75^\circ\) and \(15^\circ\) is a common question in CBSE and state board examinations.
Compound angle identities form an important conceptual base for trigonometric simplifications in JEE Main, NEET and BITSAT.
These identities are also heavily used in solving trigonometric equations, calculus limits and Fourier analysis in higher mathematics.
Students who master compound angle identities can evaluate many angles quickly without memorizing extra formulas.

Concept Used

This problem uses the compound angle identities of trigonometric functions. A key identity used in this question is the cosine addition formula.

\[ \cos A \cos B - \sin A \sin B = \cos(A+B) \]

Another identity required is the complementary angle identity

\[ \cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta \]

By applying these identities, the given expression can be simplified systematically.

Solution Roadmap

Identify the pattern \(\cos A\cos B-\sin A\sin B\).
Apply the cosine addition identity.
Simplify the resulting angle expression.
Use the complementary angle identity.
Verify that the result matches the right-hand side.

Angle Identity Visualization

Solution

We need to prove

\[ \cos \left( \dfrac{\pi}{4}-x\right)\cos \left( \dfrac{\pi}{4}-y\right) -\sin \left( \dfrac{\pi}{4}-x\right)\sin \left( \dfrac{\pi}{4}-y\right) =\sin (x+y) \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cos \left( \dfrac{\pi}{4}-x\right)\cos \left( \dfrac{\pi}{4}-y\right) -\sin \left( \dfrac{\pi}{4}-x\right)\sin \left( \dfrac{\pi}{4}-y\right) \end{aligned} \]

Using the identity

\[ \cos A\cos B-\sin A\sin B=\cos(A+B) \]

Let

\[ A=\dfrac{\pi}{4}-x, \qquad B=\dfrac{\pi}{4}-y \]

Therefore

\[ \begin{aligned} \text{LHS} &=\cos\left[\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}-y\right)\right] \end{aligned} \]

Simplifying the angle

\[ \begin{aligned} \text{LHS} &=\cos\left(\dfrac{\pi}{2}-x-y\right) \end{aligned} \]

Using the complementary identity

\[ \cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta \]

\[ \begin{aligned} \text{LHS} &=\sin(x+y) \end{aligned} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Compound angle identities are frequently tested in CBSE and state board examinations.
Problems like this help students recognize patterns such as \(\cos A\cos B-\sin A\sin B\).
Competitive examinations such as JEE Main, NEET and BITSAT require quick application of trigonometric identities to simplify expressions.
Understanding these identities is essential for solving advanced problems in trigonometric equations, calculus and Fourier analysis.

Concept Used

This problem uses the compound angle identities of the tangent function. These identities allow trigonometric expressions involving sums or differences of angles to be simplified algebraically.

Important identities used in this question are

\[ \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \]

\[ \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \]

Since \[ \tan\dfrac{\pi}{4}=1 \] the expressions simplify significantly when substituting \(A=\dfrac{\pi}{4}\).

Solution Roadmap

Apply the tangent addition formula to \(\tan(\frac{\pi}{4}+x)\).
Apply the tangent subtraction formula to \(\tan(\frac{\pi}{4}-x)\).
Substitute \(\tan\frac{\pi}{4}=1\).
Simplify both expressions.
Form the ratio and simplify algebraically.

Angle Illustration

Solution

We need to prove

\[ \dfrac{\tan \left( \dfrac{\pi}{4}+x\right)} {\tan \left( \dfrac{\pi}{4}-x\right)} = \left(\dfrac{1+\tan x}{1-\tan x}\right)^2 \]

Start with the left-hand side.

Using the tangent addition identity

\[ \tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} \]

\[ \begin{aligned} \tan\left(\dfrac{\pi}{4}+x\right) &=\dfrac{\tan\dfrac{\pi}{4}+\tan x}{1-\tan\dfrac{\pi}{4}\tan x} \end{aligned} \]

Since \(\tan\dfrac{\pi}{4}=1\),

\[ \tan\left(\dfrac{\pi}{4}+x\right) =\dfrac{1+\tan x}{1-\tan x} \]

Similarly using the tangent subtraction identity

\[ \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} \]

\[ \begin{aligned} \tan\left(\dfrac{\pi}{4}-x\right) &=\dfrac{\tan\dfrac{\pi}{4}-\tan x}{1+\tan\dfrac{\pi}{4}\tan x} \end{aligned} \]

Substituting \(\tan\dfrac{\pi}{4}=1\)

\[ \tan\left(\dfrac{\pi}{4}-x\right) =\dfrac{1-\tan x}{1+\tan x} \]

Now form the ratio.

\[ \begin{aligned} \dfrac{\tan\left(\dfrac{\pi}{4}+x\right)} {\tan\left(\dfrac{\pi}{4}-x\right)} &= \dfrac{\dfrac{1+\tan x}{1-\tan x}} {\dfrac{1-\tan x}{1+\tan x}} \end{aligned} \]

\[ \begin{aligned} &=\dfrac{(1+\tan x)^2}{(1-\tan x)^2} \end{aligned} \]

Therefore

\[ \dfrac{\tan \left( \dfrac{\pi}{4}+x\right) }{\tan \left( \dfrac{\pi}{4}-x\right) } = \left(\dfrac{1+\tan x}{1-\tan x}\right)^2 \]

Thus the identity is proved.

Significance for Board and Competitive Examinations

Compound angle identities for tangent frequently appear in CBSE and state board examinations.
Recognizing patterns such as \(\tan(\frac{\pi}{4}\pm x)\) helps students simplify trigonometric expressions quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often include problems requiring rapid use of tangent addition formulas.
These identities also play an important role in solving trigonometric equations and calculus problems involving tangent functions.

Concept Used

This identity uses allied-angle identities and properties of even and odd trigonometric functions.

Important identities used in this question are

\(\cos(\pi +x)=-\cos x\)
\(\cos(-x)=\cos x\)
\(\sin(\pi -x)=\sin x\)
\(\cos\left(\dfrac{\pi}{2}+x\right)=-\sin x\)

These identities help transform angles involving \(\pi\) and \(\dfrac{\pi}{2}\) into simpler expressions involving only \(x\).

Solution Roadmap

Apply allied-angle identities to each trigonometric term.
Convert every function into expressions involving \(x\).
Simplify numerator and denominator separately.
Express the result using basic trigonometric ratios.
Recognize the identity \(\cot x=\dfrac{\cos x}{\sin x}\).

Allied Angle Visualization

Solution

We need to prove

\[ \dfrac{\cos (\pi +x)\cos (-x)} {\sin (\pi -x)\cos \left(\dfrac{\pi}{2}+x\right)} =\cot ^2 x \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\cos (\pi +x)\cos (-x)} {\sin (\pi -x)\cos \left(\dfrac{\pi}{2}+x\right)} \end{aligned} \]

Using allied-angle identities

\[ \cos(\pi +x)=-\cos x \]

\[ \cos(-x)=\cos x \]

\[ \sin(\pi -x)=\sin x \]

\[ \cos\left(\dfrac{\pi}{2}+x\right)=-\sin x \]

Substitute these values

\[ \begin{aligned} \text{LHS} &=\dfrac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \end{aligned} \]

\[ \begin{aligned} &=\dfrac{\cos^2 x}{\sin^2 x} \end{aligned} \]

Since

\[ \cot x=\dfrac{\cos x}{\sin x} \]

we obtain

\[ \text{LHS}=\cot^2 x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Allied-angle identities are frequently tested in CBSE and other state board examinations.
Recognizing transformations like \((\pi+x)\), \((\pi-x)\), and \((\frac{\pi}{2}+x)\) helps simplify complex trigonometric expressions quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often include identities involving angle transformations.
Strong understanding of these identities is essential for solving trigonometric equations, calculus limits and advanced trigonometric proofs.

Concept Used

This problem uses allied-angle identities and periodic properties of trigonometric functions. Angles containing \(\pi\), \(\dfrac{3\pi}{2}\), or \(2\pi\) can often be simplified using known transformations.

Important identities used:

\(\cos\left(\dfrac{3\pi}{2}+x\right)=\sin x\)
\(\cos(2\pi+x)=\cos x\)
\(\cot\left(\dfrac{3\pi}{2}-x\right)=\tan x\)
\(\cot(2\pi+x)=\cot x\)
\(\sin^2x+\cos^2x=1\)

These identities allow complicated angle expressions to be rewritten in terms of simple trigonometric functions of \(x\).

Solution Roadmap

Use allied-angle identities to simplify each trigonometric term.
Rewrite all expressions in terms of \(\sin x\), \(\cos x\), \(\tan x\), and \(\cot x\).
Simplify the bracketed expression.
Use the Pythagorean identity \(\sin^2x+\cos^2x=1\).
Show that the expression reduces to 1.

Allied Angle Illustration

Solution

We need to prove

\[ \cos \left( \dfrac{3\pi}{2}+x\right) \cos \left( 2\pi +x\right) \left[ \cot \left( \dfrac{3\pi}{2}-x\right) +\cot \left( 2\pi +x\right) \right]=1 \]

Start with the left-hand side.

Using allied-angle identities

\[ \cos\left(\dfrac{3\pi}{2}+x\right)=\sin x \]

\[ \cos(2\pi+x)=\cos x \]

\[ \cot\left(\dfrac{3\pi}{2}-x\right)=\tan x \]

\[ \cot(2\pi+x)=\cot x \]

Substituting these values:

\[ \begin{aligned} \text{LHS} &= \sin x \cdot \cos x \left[ \tan x + \cot x \right] \end{aligned} \]

Express the terms using sine and cosine.

\[ \begin{aligned} \tan x &= \dfrac{\sin x}{\cos x} \ \cot x &= \dfrac{\cos x}{\sin x} \end{aligned} \]

\[ \begin{aligned} \text{LHS} &= \sin x\cos x \left[ \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} \right] \end{aligned} \]

\[ \begin{aligned} &= \sin x\cos x \left[ \dfrac{\sin^2 x+\cos^2 x}{\sin x\cos x} \right] \end{aligned} \]

Using the identity

\[ \sin^2 x+\cos^2 x=1 \]

\[ \text{LHS}=\dfrac{\sin x\cos x}{\sin x\cos x}=1 \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Allied-angle identities involving \(\pi\), \(\dfrac{3\pi}{2}\), and \(2\pi\) are frequently tested in CBSE and state board examinations.
Recognizing periodic properties of trigonometric functions allows complex expressions to be simplified quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often include identities that combine angle transformations and basic trigonometric ratios.
Strong mastery of these identities is essential for solving advanced problems in trigonometric equations, calculus, and mathematical physics.

Concept Used

This identity uses a fundamental trigonometric relation that converts products of sine and cosine functions into a cosine of angle difference.

The identity used here is

\[ \sin A \sin B + \cos A \cos B = \cos(A-B) \]

This identity is extremely useful when two trigonometric expressions contain angles that differ slightly, as in \((n+1)x\) and \((n+2)x\).

Another property used is that cosine is an even function.

\[ \cos(-\theta)=\cos\theta \]

Solution Roadmap

Recognize the pattern \(\sin A\sin B+\cos A\cos B\).
Apply the identity \(\cos(A-B)\).
Substitute \(A=(n+1)x\) and \(B=(n+2)x\).
Simplify the resulting angle expression.
Use the property \(\cos(-x)=\cos x\).

Angle Difference Illustration

Solution

We need to prove

\[ \sin (n+1)x \sin (n+2)x + \cos (n+1)x \cos (n+2)x = \cos x \]

Start with the left-hand side.

Using the identity

\[ \sin A \sin B + \cos A \cos B = \cos(A-B) \]

Let

\[ A=(n+1)x,\qquad B=(n+2)x \]

Therefore

\[ \begin{aligned} \text{LHS} &= \cos\big[(n+1)x-(n+2)x\big] \end{aligned} \]

Simplifying the angle

\[ \begin{aligned} (n+1)x-(n+2)x &= nx+x-nx-2x \ &= -x \end{aligned} \]

Thus

\[ \text{LHS}=\cos(-x) \]

Since cosine is an even function

\[ \cos(-x)=\cos x \]

Therefore

\[ \text{LHS}=\cos x=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Product identities such as \(\sin A\sin B+\cos A\cos B=\cos(A-B)\) frequently appear in CBSE and other board examinations.
These identities are extremely useful in simplifying expressions involving sequences of angles like \((n+1)x\) and \((n+2)x\).
Competitive examinations such as JEE Main, NEET and BITSAT often test recognition of such identity patterns.
Mastery of these identities helps students solve advanced trigonometric equations and calculus problems efficiently.

Concept Used

This identity uses the sum–to–product transformation for cosine functions. Such identities are useful for simplifying expressions where trigonometric functions involve sums and differences of angles.

The identity used here is

\[ \cos A - \cos B = -2\sin \left(\dfrac{A+B}{2}\right)\sin \left(\dfrac{A-B}{2}\right) \]

We also use the standard value

\[ \sin \left(\dfrac{3\pi}{4}\right)=\dfrac{1}{\sqrt{2}} \]

Solution Roadmap

Recognize the pattern \(\cos A - \cos B\).
Apply the sum–to–product identity.
Simplify the expressions \(\dfrac{A+B}{2}\) and \(\dfrac{A-B}{2}\).
Substitute the known value of \(\sin\left(\dfrac{3\pi}{4}\right)\).
Simplify the result to obtain the required identity.

Angle Illustration

Solution

We need to prove

\[ \cos \left( \dfrac{3\pi}{4}+x\right) - \cos \left( \dfrac{3\pi}{4}-x\right) = -\sqrt{2}\sin x \]

Start with the left-hand side.

Using the identity

\[ \cos A - \cos B = -2\sin\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right) \]

Let

\[ A=\dfrac{3\pi}{4}+x, \qquad B=\dfrac{3\pi}{4}-x \]

Then

\[ \dfrac{A+B}{2} = \dfrac{\left(\dfrac{3\pi}{4}+x+\dfrac{3\pi}{4}-x\right)}{2} = \dfrac{3\pi}{4} \]

and

\[ \dfrac{A-B}{2} = \dfrac{\left(\dfrac{3\pi}{4}+x-\left(\dfrac{3\pi}{4}-x\right)\right)}{2} = x \]

Therefore

\[ \begin{aligned} \text{LHS} &= -2\sin\left(\dfrac{3\pi}{4}\right)\sin x \end{aligned} \]

Using the standard value

\[ \sin\left(\dfrac{3\pi}{4}\right)=\dfrac{1}{\sqrt{2}} \]

\[ \begin{aligned} \text{LHS} &= -2\left(\dfrac{1}{\sqrt{2}}\right)\sin x \ &= -\sqrt{2}\sin x \end{aligned} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Sum–to–product identities are commonly tested in CBSE and other state board examinations.
Recognizing expressions of the form \(\cos A - \cos B\) allows students to simplify trigonometric expressions efficiently.
Competitive examinations such as JEE Main, NEET and BITSAT often include transformations between sum and product identities.
These identities are also important for solving trigonometric equations, wave equations, and Fourier analysis in advanced mathematics.

Concept Used

This identity combines algebraic factorization with sum–to–product trigonometric identities. The difference of squares formula is first applied and then converted using trigonometric sum identities.

Important identities used:

\[ a^{2}-b^{2}=(a+b)(a-b) \]

\[ \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

Also,

\[ 2\sin A\cos A=\sin 2A \]

Solution Roadmap

Use the difference of squares formula.
Apply the identities for \(\sin A+\sin B\) and \(\sin A-\sin B\).
Simplify the resulting trigonometric products.
Use the double-angle identity.
Obtain the required expression.

Angle Relationship Illustration

Solution

We need to prove

\[ \sin^{2}6x-\sin^{2}4x=\sin 2x\sin 10x \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\sin^{2}6x-\sin^{2}4x \end{aligned} \]

Using the identity \(a^2-b^2=(a+b)(a-b)\),

\[ \begin{aligned} \text{LHS} &=(\sin6x+\sin4x)(\sin6x-\sin4x) \end{aligned} \]

Apply the identities

\[ \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

Therefore

\[ \begin{aligned} \sin6x+\sin4x &=2\sin\dfrac{6x+4x}{2}\cos\dfrac{6x-4x}{2}\\ &=2\sin5x\cos x \end{aligned} \]

\[ \begin{aligned} \sin6x-\sin4x &=2\cos\dfrac{6x+4x}{2}\sin\dfrac{6x-4x}{2}\\ &=2\cos5x\sin x \end{aligned} \]

Substituting these values,

\[ \begin{aligned} \text{LHS} &=(2\sin5x\cos x)(2\cos5x\sin x) \end{aligned} \]

Rearranging,

\[ \begin{aligned} \text{LHS} &=2\sin5x\cos5x\cdot2\sin x\cos x \end{aligned} \]

Using the identity \(2\sin A\cos A=\sin2A\),

\[ \begin{aligned} \text{LHS} &=\sin10x\cdot\sin2x \end{aligned} \]

Thus

\[ \text{LHS}=\sin2x\sin10x \]

Therefore

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Problems involving \(\sin^2A-\sin^2B\) frequently appear in CBSE and other board examinations.
Recognizing factorization patterns helps simplify trigonometric expressions quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often require combining algebraic identities with trigonometric transformations.
Mastery of sum–to–product identities is essential for solving advanced trigonometric equations and calculus problems.

Concept Used

This identity combines algebraic factorization with trigonometric sum–to–product transformations. The difference of squares is first applied and then cosine identities are used to convert the expression into sine products.

Important identities used:

\[ a^2-b^2=(a+b)(a-b) \]

\[ \cos A+\cos B=2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ 2\sin A\cos A=\sin 2A \]

Solution Roadmap

Use the difference of squares identity.
Apply cosine sum and difference identities.
Simplify the resulting trigonometric expressions.
Use the double-angle identity.
Show that the result equals \(\sin4x\sin8x\).

Angle Relationship Illustration

Solution

We need to prove

\[ \cos^2 2x-\cos^2 6x=\sin4x\sin8x \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cos^2 2x-\cos^2 6x \end{aligned} \]

Using the identity \(a^2-b^2=(a+b)(a-b)\),

\[ \begin{aligned} \text{LHS} &=(\cos2x+\cos6x)(\cos2x-\cos6x) \end{aligned} \]

Apply cosine identities.

\[ \cos2x+\cos6x = 2\cos\dfrac{2x+6x}{2}\cos\dfrac{2x-6x}{2} \]

\[ =2\cos4x\cos(-2x) \]

Since \(\cos(-\theta)=\cos\theta\),

\[ =2\cos4x\cos2x \]

Similarly,

\[ \cos2x-\cos6x = -2\sin\dfrac{2x+6x}{2}\sin\dfrac{2x-6x}{2} \]

\[ =-2\sin4x\sin(-2x) \]

Since \(\sin(-\theta)=-\sin\theta\),

\[ =2\sin4x\sin2x \]

Substituting these expressions,

\[ \begin{aligned} \text{LHS} &=(2\cos4x\cos2x)(2\sin4x\sin2x) \end{aligned} \]

Rearranging,

\[ \text{LHS} = 2\cos4x\sin4x\cdot 2\cos2x\sin2x \]

Using \(2\sin A\cos A=\sin2A\),

\[ \text{LHS} = \sin8x\cdot\sin4x \]

Thus

\[ \text{LHS}=\sin4x\sin8x \]

Therefore

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving \(\cos^2A-\cos^2B\) frequently appear in CBSE and state board examinations.
Recognizing factorization patterns helps simplify trigonometric identities efficiently.
Competitive examinations such as JEE Main, NEET and BITSAT often test transformations using sum-to-product identities.
These identities are fundamental for solving trigonometric equations and calculus problems involving periodic functions.

Concept Used

This identity involves the use of sum–to–product trigonometric identities together with the double-angle identity. By combining terms appropriately, the expression can be simplified into a product form.

Important identities used:

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \cos2x=\cos^2x-\sin^2x \]

\[ 1-\sin^2x=\cos^2x \]

Solution Roadmap

Rearrange the expression so that \(\sin2x+\sin6x\) can be combined.
Apply the identity for \(\sin A+\sin B\).
Factor out the common term \(\sin4x\).
Use the double-angle identity for cosine.
Simplify the expression to obtain the required result.

Angle Relationship Illustration

Solution

We need to prove

\[ \sin2x+2\sin4x+\sin6x=4\cos^2x\sin4x \]

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\sin2x+2\sin4x+\sin6x \end{aligned} \]

Rearrange the terms

\[ \text{LHS}=\sin2x+\sin6x+2\sin4x \]

Using the identity

\[ \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin2x+\sin6x &=2\sin\dfrac{2x+6x}{2}\cos\dfrac{2x-6x}{2}\ &=2\sin4x\cos(-2x) \end{aligned} \]

Since \(\cos(-\theta)=\cos\theta\),

\[ \sin2x+\sin6x=2\sin4x\cos2x \]

Substitute this into the expression

\[ \begin{aligned} \text{LHS} &=2\sin4x\cos2x+2\sin4x \end{aligned} \]

Factor out \(2\sin4x\)

\[ \text{LHS}=2\sin4x(\cos2x+1) \]

Using

\[ \cos2x=\cos^2x-\sin^2x \]

\[ \begin{aligned} \cos2x+1 &=\cos^2x-\sin^2x+1 \end{aligned} \]

Since \(1-\sin^2x=\cos^2x\),

\[ \cos2x+1=2\cos^2x \]

Therefore

\[ \text{LHS}=2\sin4x\cdot2\cos^2x \]

\[ \text{LHS}=4\cos^2x\sin4x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving multiple sine terms are frequently simplified using sum-to-product identities in CBSE and state board examinations.
Recognizing opportunities to combine terms such as \(\sin2x+\sin6x\) helps simplify trigonometric expressions efficiently.
Competitive examinations such as JEE Main, NEET and BITSAT often test the ability to manipulate trigonometric identities involving multiple angles.
These techniques are essential for solving trigonometric equations and advanced calculus problems involving periodic functions.

Concept Used

This identity uses the sum–to–product transformations for sine functions along with the definition of the cotangent function. By converting the sine sums and differences into products, both sides of the identity can be simplified to the same expression.

Important identities used:

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \cot \theta=\dfrac{\cos\theta}{\sin\theta} \]

Solution Roadmap

Apply the identity for \(\sin A+\sin B\) in the left-hand side.
Use the definition of cotangent to simplify the expression.
Apply the identity for \(\sin A-\sin B\) in the right-hand side.
Simplify both expressions to obtain the same result.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cot4x(\sin5x+\sin3x) \end{aligned} \]

Using the identity

\[ \sin A+\sin B=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin5x+\sin3x &=2\sin\dfrac{5x+3x}{2}\cos\dfrac{5x-3x}{2}\\ &=2\sin4x\cos x \end{aligned} \]

Substituting this,

\[ \begin{aligned} \text{LHS} &=\cot4x(2\sin4x\cos x) \end{aligned} \]

Using \(\cot4x=\dfrac{\cos4x}{\sin4x}\),

\[ \begin{aligned} \text{LHS} &=\dfrac{\cos4x}{\sin4x}\cdot2\sin4x\cos x\\ &=2\cos4x\cos x \end{aligned} \]

Now consider the right-hand side.

\[ \begin{aligned} \text{RHS} &=\cot x(\sin5x-\sin3x) \end{aligned} \]

Using the identity

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin5x-\sin3x &=2\cos\dfrac{5x+3x}{2}\sin\dfrac{5x-3x}{2}\\ &=2\cos4x\sin x \end{aligned} \]

Substituting this,

\[ \begin{aligned} \text{RHS} &=\cot x(2\cos4x\sin x) \end{aligned} \]

Using \(\cot x=\dfrac{\cos x}{\sin x}\),

\[ \begin{aligned} \text{RHS} &=\dfrac{\cos x}{\sin x}\cdot2\cos4x\sin x\\ &=2\cos4x\cos x \end{aligned} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Identities involving sums and differences of sine functions are frequently tested in CBSE and state board examinations.
Recognizing patterns such as \(\sin A+\sin B\) and \(\sin A-\sin B\) allows complex expressions to be simplified quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often require efficient manipulation of trigonometric identities involving multiple angles.
These transformations are also useful in solving trigonometric equations and advanced calculus problems.

Concept Used

This identity uses the sum–to–product transformations for trigonometric functions. Expressions involving differences of cosine and sine functions can be simplified using standard identities.

Important identities used:

\[ \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

These identities help convert complex trigonometric expressions into simpler products.

Solution Roadmap

Apply the identity for \(\cos A-\cos B\) in the numerator.
Apply the identity for \(\sin A-\sin B\) in the denominator.
Simplify the resulting expressions.
Cancel common factors.
Obtain the required ratio.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\cos9x-\cos5x}{\sin17x-\sin3x} \end{aligned} \]

Using the identity

\[ \cos A-\cos B=-2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \begin{aligned} \cos9x-\cos5x &=-2\sin\dfrac{9x+5x}{2}\sin\dfrac{9x-5x}{2}\\ &=-2\sin7x\sin2x \end{aligned} \]

Now apply the identity

\[ \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin17x-\sin3x &=2\cos\dfrac{17x+3x}{2}\sin\dfrac{17x-3x}{2}\\ &=2\cos10x\sin7x \end{aligned} \]

Substitute these expressions into the fraction.

\[ \begin{aligned} \text{LHS} &=\dfrac{-2\sin7x\sin2x}{2\cos10x\sin7x} \end{aligned} \]

Cancel the common factor \(\sin7x\).

\[ \text{LHS}=\dfrac{\sin2x}{\cos10x} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving \(\cos A-\cos B\) and \(\sin A-\sin B\) are common in CBSE and state board examinations.
Recognizing sum–to–product transformations helps simplify complicated trigonometric expressions quickly.
Competitive examinations such as JEE Main, NEET and BITSAT frequently test such transformations involving multiple angles.
These techniques are also important in solving trigonometric equations and advanced calculus problems involving periodic functions.

Concept Used

This identity is simplified using the sum–to–product transformations for sine and cosine functions. These identities convert sums of trigonometric functions into products, making simplification easier.

Important identities used:

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

Also,

\[ \tan\theta=\dfrac{\sin\theta}{\cos\theta} \]

Solution Roadmap

Apply the identity for \(\sin A+\sin B\) in the numerator.
Apply the identity for \(\cos A+\cos B\) in the denominator.
Simplify the resulting expression.
Cancel common factors.
Recognize the tangent ratio.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\sin5x+\sin3x}{\cos5x+\cos3x} \end{aligned} \]

Using the identity

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin5x+\sin3x &=2\sin\dfrac{5x+3x}{2}\cos\dfrac{5x-3x}{2}\\ &=2\sin4x\cos x \end{aligned} \]

Similarly, using

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \cos5x+\cos3x &=2\cos\dfrac{5x+3x}{2}\cos\dfrac{5x-3x}{2}\\ &=2\cos4x\cos x \end{aligned} \]

Substituting these expressions,

\[ \begin{aligned} \text{LHS} &=\dfrac{2\sin4x\cos x}{2\cos4x\cos x} \end{aligned} \]

Cancel the common factors \(2\) and \(\cos x\).

\[ \text{LHS}=\dfrac{\sin4x}{\cos4x} \]

Therefore

\[ \text{LHS}=\tan4x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving sums of sine and cosine functions are frequently simplified using sum-to-product identities in CBSE and state board examinations.
Recognizing patterns such as \(\sin A+\sin B\) and \(\cos A+\cos B\) helps simplify trigonometric ratios quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often include similar problems requiring efficient use of trigonometric identities.
These transformations are fundamental for solving trigonometric equations and advanced problems involving periodic functions.

Concept Used

This identity is simplified using the sum–to–product transformations for sine and cosine functions. These identities allow expressions involving sums or differences of trigonometric functions to be rewritten as products.

Important identities used:

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

Also,

\[ \tan\theta=\dfrac{\sin\theta}{\cos\theta} \]

Solution Roadmap

Apply the identity for \(\sin A-\sin B\) in the numerator.
Apply the identity for \(\cos A+\cos B\) in the denominator.
Simplify the resulting expression.
Cancel common factors.
Recognize the tangent ratio.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\sin x-\sin y}{\cos x+\cos y} \end{aligned} \]

Using the identity

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin x-\sin y &=2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2} \end{aligned} \]

Similarly, using

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \cos x+\cos y &=2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2} \end{aligned} \]

Substituting these expressions,

\[ \begin{aligned} \text{LHS} &=\dfrac{2\cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}} {2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}} \end{aligned} \]

Cancel the common factors \(2\cos\dfrac{x+y}{2}\).

\[ \text{LHS} = \dfrac{\sin\dfrac{x-y}{2}} {\cos\dfrac{x-y}{2}} \]

Therefore

\[ \text{LHS} = \tan\dfrac{x-y}{2} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Identities involving \(\sin A-\sin B\) and \(\cos A+\cos B\) are commonly tested in CBSE and other board examinations.
Recognizing these patterns helps simplify complicated trigonometric ratios quickly.
Competitive examinations such as JEE Main, NEET and BITSAT frequently require efficient use of sum–to–product identities.
These transformations are also useful in solving trigonometric equations and advanced problems involving periodic functions.

Concept Used

This identity is simplified using the sum–to–product transformations for sine and cosine functions. These identities convert sums of trigonometric functions into products, which often reveal simple ratios.

Important identities used:

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

Also,

\[ \tan\theta=\dfrac{\sin\theta}{\cos\theta} \]

Solution Roadmap

Apply the identity for \(\sin A+\sin B\) in the numerator.
Apply the identity for \(\cos A+\cos B\) in the denominator.
Simplify the resulting expression.
Cancel common factors.
Recognize the tangent ratio.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\sin x+\sin3x}{\cos x+\cos3x} \end{aligned} \]

Using the identity

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin x+\sin3x &=2\sin\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}\\ &=2\sin2x\cos(-x) \end{aligned} \]

Since \(\cos(-x)=\cos x\),

\[ \sin x+\sin3x=2\sin2x\cos x \]

Similarly, using

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \cos x+\cos3x &=2\cos\dfrac{x+3x}{2}\cos\dfrac{x-3x}{2}\\ &=2\cos2x\cos(-x) \end{aligned} \]

Since \(\cos(-x)=\cos x\),

\[ \cos x+\cos3x=2\cos2x\cos x \]

Substituting these expressions,

\[ \begin{aligned} \text{LHS} &=\dfrac{2\sin2x\cos x}{2\cos2x\cos x} \end{aligned} \]

Cancel the common factors \(2\) and \(\cos x\).

\[ \text{LHS} = \dfrac{\sin2x}{\cos2x} \]

Therefore

\[ \text{LHS} = \tan2x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving \(\sin A+\sin B\) and \(\cos A+\cos B\) are frequently tested in CBSE and other board examinations.
Recognizing average-angle patterns helps simplify trigonometric expressions quickly.
Competitive examinations such as JEE Main, NEET and BITSAT often require efficient manipulation of trigonometric identities.
These transformations are fundamental in solving trigonometric equations and advanced problems involving periodic functions.

Concept Used

This identity combines sum–to–product transformations for sine functions with double-angle identities. The expression in the numerator is simplified using trigonometric identities, while the denominator is rewritten using a cosine double-angle relation.

Important identities used:

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \cos 2x=\cos^{2}x-\sin^{2}x \]

Hence,

\[ \sin^{2}x-\cos^{2}x=-\cos2x \]

Solution Roadmap

Apply the identity for \(\sin A-\sin B\) in the numerator.
Rewrite the denominator using the double-angle identity.
Simplify the trigonometric expressions.
Cancel common factors.
Obtain the required result.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\sin x-\sin 3x}{\sin^{2}x-\cos^{2}x} \end{aligned} \]

Using the identity

\[ \sin A-\sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin x-\sin 3x &=2\cos\dfrac{x+3x}{2}\sin\dfrac{x-3x}{2}\\ &=2\cos2x\sin(-x) \end{aligned} \]

Since \(\sin(-x)=-\sin x\),

\[ \sin x-\sin 3x=-2\cos2x\sin x \]

Now simplify the denominator.

\[ \sin^{2}x-\cos^{2}x=-\cos2x \]

Substitute these expressions.

\[ \begin{aligned} \text{LHS} &=\dfrac{-2\cos2x\sin x}{-\cos2x} \end{aligned} \]

Cancel the common factor \(\cos2x\).

\[ \text{LHS}=2\sin x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving \(\sin A-\sin B\) frequently appear in CBSE and other board examinations.
Recognizing that \(\sin^{2}x-\cos^{2}x=-\cos2x\) helps simplify many trigonometric identities.
Competitive examinations such as JEE Main, NEET and BITSAT often test the ability to combine algebraic and trigonometric identities.
These techniques are essential for solving trigonometric equations and advanced calculus problems involving periodic functions.

Concept Used

This identity is simplified using sum–to–product transformations for sine and cosine functions. By combining pairs of trigonometric terms, the expression can be factorized to reveal a simple trigonometric ratio.

Important identities used:

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

Also,

\[ \cot\theta=\dfrac{\cos\theta}{\sin\theta} \]

Solution Roadmap

Group the terms \(\cos4x+\cos2x\) and \(\sin4x+\sin2x\).
Apply sum–to–product identities.
Factor common terms.
Cancel the common factor.
Recognize the cotangent ratio.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\dfrac{\cos4x+\cos3x+\cos2x}{\sin4x+\sin3x+\sin2x} \end{aligned} \]

Group the first and third terms.

\[ \text{LHS} = \dfrac{(\cos4x+\cos2x)+\cos3x}{(\sin4x+\sin2x)+\sin3x} \]

Using the identity

\[ \cos A+\cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \cos4x+\cos2x &=2\cos\dfrac{4x+2x}{2}\cos\dfrac{4x-2x}{2}\ &=2\cos3x\cos x \end{aligned} \]

Similarly,

\[ \sin A+\sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \]

\[ \begin{aligned} \sin4x+\sin2x &=2\sin\dfrac{4x+2x}{2}\cos\dfrac{4x-2x}{2}\\ &=2\sin3x\cos x \end{aligned} \]

Substitute these expressions.

\[ \begin{aligned} \text{LHS} &=\dfrac{2\cos3x\cos x+\cos3x}{2\sin3x\cos x+\sin3x} \end{aligned} \]

Factor the common terms.

\[ \text{LHS} = \dfrac{\cos3x(2\cos x+1)}{\sin3x(2\cos x+1)} \]

Cancel the common factor \((2\cos x+1)\).

\[ \text{LHS} = \dfrac{\cos3x}{\sin3x} \]

Therefore

\[ \text{LHS}=\cot3x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Expressions involving three trigonometric terms often simplify by grouping terms and applying sum–to–product identities.
Such manipulations are commonly tested in CBSE and state board examinations.
Competitive examinations such as JEE Main, NEET and BITSAT frequently require recognizing hidden patterns in trigonometric expressions.
Mastery of these identities helps students simplify complex trigonometric equations and expressions efficiently.

Concept Used

This identity involves algebraic manipulation of cotangent functions and the use of identities connecting cotangent values of different angles. The relationship between cotangent values can be derived from trigonometric identities involving sine and cosine.

An important identity used here is

\[ \cot (A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A} \]

This identity allows expressions containing products of cotangent functions to be simplified.

Solution Roadmap

Group terms to factor a common cotangent expression.
Use the cotangent difference identity.
Simplify the resulting expression.
Cancel common factors.
Show that the expression reduces to 1.

Angle Relationship Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cot x\cot2x-\cot2x\cot3x-\cot3x\cot x \end{aligned} \]

Group the first two terms.

\[ \text{LHS} = \cot x(\cot2x-\cot3x)-\cot2x\cot3x \]

Using the identity

\[ \cot(A-B)=\dfrac{\cot A\cot B+1}{\cot B-\cot A} \]

Rearranging gives

\[ \cot A-\cot B=\dfrac{\cot A\cot B+1}{\cot(A-B)} \]

Let \(A=2x\) and \(B=3x\).

\[ \cot2x-\cot3x=\dfrac{\cot2x\cot3x+1}{\cot(2x-3x)} \]

Since \(2x-3x=-x\),

\[ \cot(2x-3x)=\cot(-x)=-\cot x \]

Substitute this into the expression.

\[ \begin{aligned} \text{LHS} &=\cot x\left(\dfrac{\cot2x\cot3x+1}{-\cot x}\right)-\cot2x\cot3x \end{aligned} \]

Simplify.

\[ \begin{aligned} \text{LHS} &=-(\cot2x\cot3x+1)+\cot2x\cot3x \end{aligned} \]

\[ \text{LHS}=1 \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Identities involving products of cotangent functions are frequently tested in CBSE and other board examinations.
Recognizing patterns and grouping terms is an important technique for simplifying trigonometric expressions.
Competitive examinations such as JEE Main, NEET and BITSAT often include identities requiring algebraic manipulation of trigonometric ratios.
Such identities strengthen conceptual understanding required for solving advanced trigonometric equations and calculus problems.

Concept Used

This identity is derived using the multiple-angle identities of the tangent function. The angle \(4x\) is expressed as \(2(2x)\), and the tangent double-angle identity is applied repeatedly.

x Important identities used:

\[ \tan 2\theta=\dfrac{2\tan\theta}{1-\tan^{2}\theta} \]

and

\[ \tan 2A=\dfrac{2\tan A}{1-\tan^{2}A} \]

These identities allow higher-angle tangent expressions to be written entirely in terms of \(\tan x\).

Solution Roadmap

Write \(4x\) as \(2(2x)\).
Apply the tangent double-angle identity.
Substitute the formula for \(\tan 2x\).
Simplify the resulting algebraic expression.
Express the final result entirely in terms of \(\tan x\).

Angle Progression Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\tan4x \end{aligned} \]

Write \(4x=2(2x)\).

\[ \tan4x=\tan(2\cdot2x) \]

Using the double-angle identity

\[ \tan2A=\dfrac{2\tan A}{1-\tan^{2}A} \]

\[ \begin{aligned} \tan4x &=\dfrac{2\tan2x}{1-\tan^{2}2x} \end{aligned} \]

Now substitute

\[ \tan2x=\dfrac{2\tan x}{1-\tan^{2}x} \]

\[ \begin{aligned} \tan4x &=\dfrac{2\left(\dfrac{2\tan x}{1-\tan^{2}x}\right)}{1-\tan^{2}2x} \end{aligned} \]

\[ \tan4x=\dfrac{4\tan x}{(1-\tan^{2}x)(1-\tan^{2}2x)} \]

Now evaluate \(1-\tan^{2}2x\).

\[ 1-\tan^{2}2x = (1+\tan2x)(1-\tan2x) \]

Substitute

\[ \tan2x=\dfrac{2\tan x}{1-\tan^{2}x} \]

\[ \begin{aligned} \tan4x &=\dfrac{4\tan x}{(1-\tan^{2}x) \left(1+\dfrac{2\tan x}{1-\tan^{2}x}\right) \left(1-\dfrac{2\tan x}{1-\tan^{2}x}\right)} \end{aligned} \]

Simplify the denominator.

\[ (1-\tan^{2}x+2\tan x)(1-\tan^{2}x-2\tan x) \]

\[ =(1-\tan^{2}x)^{2}-(2\tan x)^{2} \]

\[ =1-6\tan^{2}x+\tan^{4}x \]

Therefore

\[ \tan4x=\dfrac{4\tan x(1-\tan^{2}x)}{1-6\tan^{2}x+\tan^{4}x} \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Multiple-angle identities such as \(\tan4x\) expressed in terms of \(\tan x\) are important in CBSE and state board examinations.
Such identities are commonly used in trigonometric simplifications and equation solving.
Competitive examinations such as JEE Main, NEET and BITSAT frequently test derivations involving double-angle and multiple-angle formulas.
Understanding these transformations helps students tackle advanced problems in trigonometry and calculus involving periodic functions.

Concept Used

This identity is derived using the double-angle formulas for cosine and sine. The angle \(4x\) is expressed as \(2(2x)\), and the cosine double-angle identity is applied.

Important identities used:

\[ \cos 2\theta=\cos^2\theta-\sin^2\theta \]

\[ \cos 2\theta=1-2\sin^2\theta \]

and

\[ \sin2x=2\sin x\cos x \]

These identities allow the expression \(\cos4x\) to be written entirely in terms of \(\sin x\) and \(\cos x\).

Solution Roadmap

Write \(4x\) as \(2(2x)\).
Apply the cosine double-angle identity.
Rewrite the expression using \(\sin^2\) identities.
Substitute the value of \(\sin2x\).
Simplify to obtain the required expression.

Angle Doubling Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cos4x \end{aligned} \]

Write \(4x=2(2x)\).

\[ \cos4x=\cos2(2x) \]

Using the double-angle identity

\[ \cos2A=\cos^2A-\sin^2A \]

\[ \begin{aligned} \cos4x &=\cos^22x-\sin^22x \end{aligned} \]

Using the identity \(\cos^2\theta=1-\sin^2\theta\),

\[ \begin{aligned} \cos4x &=(1-\sin^22x)-\sin^22x \end{aligned} \]

\[ \cos4x=1-2\sin^22x \]

Now use the identity

\[ \sin2x=2\sin x\cos x \]

\[ \sin^22x=(2\sin x\cos x)^2 \]

\[ \sin^22x=4\sin^2x\cos^2x \]

Substitute this value.

\[ \begin{aligned} \cos4x &=1-2(4\sin^2x\cos^2x) \end{aligned} \]

\[ \cos4x=1-8\sin^2x\cos^2x \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Multiple-angle identities such as \(\cos4x\) expressed in terms of \(\sin x\) and \(\cos x\) are frequently asked in CBSE and other board examinations.
These identities help simplify trigonometric expressions and equations efficiently.
Competitive examinations such as JEE Main, NEET and BITSAT often test derivations involving double-angle and multiple-angle formulas.
Understanding these identities is essential for solving advanced trigonometric and calculus problems involving periodic functions.

Concept Used

This identity is derived using the multiple-angle identities of cosine. The angle \(6x\) is written as \(3(2x)\), allowing the triple-angle identity of cosine to be applied.

Important identities used:

\[ \cos 3\theta = 4\cos^{3}\theta - 3\cos\theta \]

\[ \cos 2x = 2\cos^{2}x - 1 \]

\[ \sin 2x = 2\sin x\cos x \]

These identities allow the expression \(\cos 6x\) to be rewritten entirely in terms of powers of \(\cos x\).

Solution Roadmap

Write \(6x = 3(2x)\).
Apply the cosine triple-angle identity.
Express \(\cos 2x\) and \(\sin 2x\) in terms of \(\sin x\) and \(\cos x\).
Convert all expressions into powers of \(\cos x\).
Simplify the algebraic expression.

Angle Transformation Illustration

Solution

Start with the left-hand side.

\[ \begin{aligned} \text{LHS} &=\cos6x \end{aligned} \]

Write \(6x = 3(2x)\).

\[ \cos6x=\cos3(2x) \]

Using the identity

\[ \cos3\theta=4\cos^{3}\theta-3\cos\theta \]

\[ \begin{aligned} \cos6x &=4\cos^{3}2x-3\cos2x \end{aligned} \]

Factor \(\cos2x\).

\[ \cos6x=\cos2x(4\cos^{2}2x-3) \]

Now write \(\cos^{2}2x\) in terms of sine.

\[ \cos^{2}2x=1-\sin^{2}2x \]

\[ \sin2x=2\sin x\cos x \]

\[ \sin^{2}2x=4\sin^{2}x\cos^{2}x \]

Substitute these values.

\[ \begin{aligned} \cos6x &=\cos2x\left[4(1-4\sin^{2}x\cos^{2}x)-3\right] \end{aligned} \]

\[ \cos6x=\cos2x(1-16\sin^{2}x\cos^{2}x) \]

Using

\[ \cos2x=2\cos^{2}x-1 \]

and

\[ \sin^{2}x=1-\cos^{2}x \]

\[ \begin{aligned} \cos6x &=(2\cos^{2}x-1)\left[1-16(1-\cos^{2}x)\cos^{2}x\right] \end{aligned} \]

Simplify the expression.

\[ (2\cos^{2}x-1)(1-16\cos^{2}x+16\cos^{4}x) \]

Expanding,

\[ 32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 \]

Thus

\[ \text{LHS}=\text{RHS} \]

Hence the identity is proved.

Significance for Board and Competitive Examinations

Multiple-angle identities such as \(\cos 6x\) expressed in powers of \(\cos x\) are frequently asked in CBSE and state board examinations.
These identities help simplify trigonometric expressions involving higher multiples of angles.
Competitive examinations such as JEE Main, NEET and BITSAT often test derivations using triple-angle and double-angle identities.
Understanding such transformations is essential for solving advanced trigonometric and calculus problems involving periodic functions.

Chapter 3 — TRIGONOMETRIC FUNCTIONS

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Illustration of Standard Angles

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Quadrant Illustration for \(7\pi/6\)

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Unit Circle Illustration for \(30^\circ\) and \(150^\circ\)

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Unit Circle Illustration of \(45^\circ\), \(60^\circ\), and \(135^\circ\)

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Angle Combination Illustration

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Angle Identity Visualization

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Angle Illustration

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Allied Angle Visualization

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Allied Angle Illustration

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Angle Difference Illustration

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Angle Illustration

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Angle Relationship Illustration

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Angle Relationship Illustration

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Angle Relationship Illustration

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Angle Relationship Illustration

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Angle Relationship Illustration

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