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Class 10 Mathematics Exercise 8.3 NCERT Solutions Olympiad Board Exam

Chapter 8 — Introduction to Trigonometry

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

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Q1
NUMERIC3 marks

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Concept Used (Theory)

  • Basic trigonometric identities: \[\begin{aligned}\cot A&=\dfrac{\cos A}{\sin A}, \\\\ \tan A&=\dfrac{1}{\cot A}\end{aligned}\]
  • Pythagorean identity: \[\sin^2 A+\cos^2 A=1\]
  • Derived identities: \[\begin{aligned}1+\cot^2 A&=\text{cosec }^2 A \quad \text{and} \\ 1+\tan^2 A&=\sec^2 A\end{aligned}\]
  • All ratios can be expressed in terms of a single ratio using identities.

Solution Roadmap

  • Start from the given ratio: cot A
  • Use identity: $\cot A = \dfrac{\cos A}{\sin A}$
  • Express one ratio in terms of the other
  • Substitute into $\sin^2 A + \cos^2 A = 1$
  • Solve algebraically step-by-step
Base (Adjacent) Height (Opposite) Hypotenuse A

Solution

1. sin A in terms of cot A
\[\begin{aligned} \cot A &= \dfrac{\cos A}{\sin A} \\ \Rightarrow \cos A &= \cot A \cdot \sin A \\ \text{Using identity: } \sin^2 A + \cos^2 A &= 1 \\ \sin^2 A + (\cot A \cdot \sin A)^2 &= 1 \\ \sin^2 A + \cot^2 A \sin^2 A &= 1 \\ \sin^2 A (1+\cot^2 A) &= 1 \\ \sin^2 A &= \dfrac{1}{1+\cot^2 A} \\ \sin A &= \dfrac{1}{\sqrt{1+\cot^2 A}} \end{aligned}\]
2. sec A in terms of cot A
\[\begin{aligned} \sin A &= \dfrac{1}{\sqrt{1+\cot^2 A}} \\ \text{From } \cot A &= \dfrac{\cos A}{\sin A} \\\Rightarrow \cos A &= \cot A \cdot \sin A \\ \cos A &= \cot A \cdot \dfrac{1}{\sqrt{1+\cot^2 A}} \\ &= \dfrac{\cot A}{\sqrt{1+\cot^2 A}} \\ \text{Now } \sec A &= \dfrac{1}{\cos A} \\ \sec A &= \dfrac{\sqrt{1+\cot^2 A}}{\cot A} \end{aligned}\]
3. tan A in terms of cot A
\[\begin{aligned} \tan A = \dfrac{1}{\cot A} \end{aligned}\]

Final Results

  • \[\sin A = \dfrac{1}{\sqrt{1+\cot^2 A}}\]
  • \[\sec A = \dfrac{\sqrt{1+\cot^2 A}}{\cot A}\]
  • \[\tan A = \dfrac{1}{\cot A}\]

Exam Significance

  • Very important for CBSE Board Exams: conversion of ratios is frequently asked.
  • Builds foundation for proving identities in higher exercises.
  • Common in competitive exams like JEE, NTSE, and Olympiads.
  • Helps in solving complex trigonometric simplifications quickly.
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1 / 4  ·  25%
Q2 →
Q2
NUMERIC3 marks

Write all the other trigonometric ratios of \(\angle) A in terms of sec A.

Concept Used (Theory)

  • Reciprocal identity: \[\sec A = \dfrac{1}{\cos A}\]
  • Pythagorean identity: \[\sin^2 A + \cos^2 A = 1\]
  • Derived identity: \[\sec^2 A - 1 = \tan^2 A\]
  • Using these identities, all trigonometric ratios can be expressed in terms of sec A.

Solution Roadmap

  • Start with \(\sec A = \dfrac{1}{\cos A}\) to find cos A
  • Use identity \(\sin^2 A + \cos^2 A = 1\) to find sin A
  • Find cosec A using reciprocal of sin A
  • Use \(\tan A = \dfrac{\sin A}{\cos A}\)
  • Find cot A as reciprocal of tan A
Adjacent Opposite Hypotenuse A

Solution

1. cos A in terms of sec A
\[ \begin{aligned} \sec A &= \dfrac{1}{\cos A} \\ \Rightarrow \cos A &= \dfrac{1}{\sec A} \end{aligned} \]
2. sin A and cosec A in terms of sec A
\[ \begin{aligned} \text{Using identity: } \sin^2 A + \cos^2 A &= 1 \\ \sin^2 A + \left(\dfrac{1}{\sec A}\right)^2 &= 1 \\ \sin^2 A + \dfrac{1}{\sec^2 A} &= 1 \\ \sin^2 A &= 1 - \dfrac{1}{\sec^2 A} \\ \sin^2 A &= \dfrac{\sec^2 A - 1}{\sec^2 A} \\ \sin A &= \dfrac{\sqrt{\sec^2 A - 1}}{\sec A} \end{aligned} \] \[ \begin{aligned} \text{cosec } A &= \dfrac{1}{\sin A} \\ \text{cosec } A &= \dfrac{\sec A}{\sqrt{\sec^2 A - 1}} \end{aligned} \]
3. tan A and cot A in terms of sec A
\[ \begin{aligned} \tan A &= \dfrac{\sin A}{\cos A} \\ \tan A &= \dfrac{\dfrac{\sqrt{\sec^2 A - 1}}{\sec A}}{\dfrac{1}{\sec A}} \\ \tan A &= \dfrac{\sqrt{\sec^2 A - 1}}{\sec A} \times \dfrac{\sec A}{1} \\ \tan A &= \sqrt{\sec^2 A - 1} \end{aligned} \] \[ \begin{aligned} \cot A &= \dfrac{1}{\tan A} \\ \cot A &= \dfrac{1}{\sqrt{\sec^2 A - 1}} \end{aligned} \]

Final Results

  • \[\cos A = \dfrac{1}{\sec A}\]
  • \[\sin A = \dfrac{\sqrt{\sec^2 A - 1}}{\sec A}\]
  • \[\text{cosec } A = \dfrac{\sec A}{\sqrt{\sec^2 A - 1}}\]
  • \[\tan A = \sqrt{\sec^2 A - 1}\]
  • \[\cot A = \dfrac{1}{\sqrt{\sec^2 A - 1}}\]

Exam Significance

  • Important for CBSE Board Exams: conversion between ratios is a standard question.
  • Directly used in proving trigonometric identities.
  • Highly useful in competitive exams like JEE, NTSE, and Olympiads.
  • Strengthens algebraic manipulation skills in trigonometry.
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Q3 →
Q3
NUMERIC3 marks

Choose the correct option. Justify your choice.

Concept Used (Theory)

  • Pythagorean identities: \[\sec^2 A - \tan^2 A = 1\] \[1 + \tan^2 A = \sec^2 A\] \[1 + \cot^2 A = \text{cosec }^2 A\]
  • Algebraic identities: \[(a+b)(a-b) = a^2 - b^2\]
  • Basic ratios: \[\tan A = \dfrac{\sin A}{\cos A}, \quad \sec A = \dfrac{1}{\cos A}\]

Solution Roadmap

  • Convert expressions into basic sine and cosine where required
  • Apply standard identities
  • Simplify step-by-step without skipping algebra
  • Match final result with given options

Solution

i.

\[ \begin{aligned} 9\sec^2 A - 9\tan^2 A \\ = 9(\sec^2 A - \tan^2 A) \\ = 9(1) \quad \text{[since } \sec^2 A - \tan^2 A = 1] \\ = 9 \end{aligned} \]
(B) 9 is correct option

ii.

\[ \begin{aligned} (1+\tan \theta +\sec \theta)(1+\cot \theta - \text{cosec } \theta) \\ = \left(1+\dfrac{\sin \theta}{\cos \theta}+\dfrac{1}{\cos \theta}\right) \left(1+\dfrac{\cos \theta}{\sin \theta}-\dfrac{1}{\sin \theta}\right) \\ = \left(\dfrac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\dfrac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \\ = \dfrac{(\sin \theta + \cos \theta + 1)(\sin \theta + \cos \theta - 1)}{\sin \theta \cos \theta} \end{aligned} \]

Let \(a = \sin \theta + \cos \theta\), then

\[ \begin{aligned} = \dfrac{a^2 - 1}{\sin \theta \cos \theta} \\ = \dfrac{(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \\ = \dfrac{(1 + 2\sin \theta \cos \theta - 1)}{\sin \theta \cos \theta} \\ = \dfrac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} \\ = 2 \end{aligned} \]
(C) 2 is correct option

iii.

\[ \begin{aligned} (\sec A + \tan A)(1 - \sin A) \\ = \left(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\right)(1 - \sin A) \\ = \dfrac{1 + \sin A}{\cos A}(1 - \sin A) \\ = \dfrac{(1+\sin A)(1-\sin A)}{\cos A} \\ = \dfrac{1 - \sin^2 A}{\cos A} \\ = \dfrac{\cos^2 A}{\cos A} \\ = \cos A \end{aligned} \]
(D) cos A is correct option

iv.

\[ \begin{aligned} \dfrac{1+\tan^2 A}{1+\cot^2 A} \\ = \dfrac{\sec^2 A}{\text{cosec }^2 A} \\ = \dfrac{\dfrac{1}{\cos^2 A}}{\dfrac{1}{\sin^2 A}} \\ = \dfrac{1}{\cos^2 A} \times \dfrac{\sin^2 A}{1} \\ = \dfrac{\sin^2 A}{\cos^2 A} \\ = \tan^2 A \end{aligned} \]
(D) tan² A is correct option

Exam Significance

  • Very important for CBSE Board Exams: MCQs and simplification-based questions are common.
  • Strengthens identity recognition and algebraic manipulation.
  • Frequently appears in competitive exams like JEE, NTSE, and Olympiads.
  • Helps in solving higher-level trigonometric equations quickly.
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3 / 4  ·  75%
Q4 →
Q4
NUMERIC3 marks

Prove the following identities (with complete logic and explanation).

Core Logic to Remember

  • Always start from LHS unless RHS is simpler.
  • Convert everything into \(\sin\) and \(\cos\).
  • Use identities only when required, not randomly.
  • Factorization is the key step in most proofs.
  • Cancel only after proper factorization.

Solution with Explanation

i.

\[ (\text{cosec } \theta - \cot \theta)^2 \] \[ \begin{aligned} &=\left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 && \text{(convert into sin, cos)}\\ &=\left(\frac{1-\cos \theta}{\sin \theta}\right)^2 && \text{(take common denominator)}\\ &=\frac{(1-\cos \theta)^2}{\sin^2 \theta} && \text{(square numerator and denominator)}\\ &=\frac{(1-\cos \theta)^2}{1-\cos^2 \theta} && \text{(use } \sin^2\theta = 1-\cos^2\theta\text{)}\\ &=\frac{(1-\cos \theta)^2}{(1-\cos \theta)(1+\cos \theta)} && \text{(apply } a^2-b^2\text{)}\\ &=\frac{1-\cos \theta}{1+\cos \theta} && \text{(cancel common factor)} \end{aligned} \]

ii.

\[ \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A} \] \[ \begin{aligned} &=\frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)\cos A} && \text{(take LCM)}\\ &=\frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{(1+\sin A)\cos A} && \text{(expand square)}\\ &=\frac{1 + 1 + 2\sin A}{(1+\sin A)\cos A} && \text{(use } \sin^2A+\cos^2A=1\text{)}\\ &=\frac{2(1+\sin A)}{(1+\sin A)\cos A} && \text{(factorize)}\\ &=\frac{2}{\cos A} = 2\sec A && \text{(cancel and use reciprocal)} \end{aligned} \]

iii.

\[ \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} \] \[ \begin{aligned} &=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} +\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}} && \text{(convert ratios)}\\ &=\frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} +\frac{\cos^2 \theta}{\sin \theta(\cos \theta-\sin \theta)} && \text{(simplify denominators)}\\ &=\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta-\cos \theta)} && \text{(combine fractions)}\\ &=\frac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta \cos \theta)} {\sin \theta \cos \theta (\sin \theta-\cos \theta)} && \text{(use } a^3-b^3\text{)}\\ &=\frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta} && \text{(cancel factor and use identity)}\\ &=\frac{1}{\sin \theta \cos \theta}+1 && \text{(split fraction)}\\ &=\sec \theta \text{cosec } \theta + 1 && \text{(convert to ratios)} \end{aligned} \]

iv.

LHS

\[ \begin{aligned} \frac{1+\sec A}{\sec A} &=\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}} && \text{(convert sec)}\\ &=\frac{\cos A+1}{1} && \text{(simplify)}\\ &=1+\cos A \end{aligned} \]

RHS

\[ \begin{aligned} \frac{\sin^2 A}{1-\cos A} &=\frac{1-\cos^2 A}{1-\cos A} && \text{(use identity)}\\ &=\frac{(1-\cos A)(1+\cos A)}{1-\cos A} && \text{(factorize)}\\ &=1+\cos A && \text{(cancel)} \end{aligned} \]

v.

\[ \frac{\cos A-\sin A+1}{\cos A+\sin A-1} \] \[ \begin{aligned} &=\frac{\cot A-1+\text{cosec } A}{\cot A+1-\text{cosec } A} && \text{(divide by sin A)}\\ &=\frac{\cot A-(1-\text{cosec } A)}{\cot A+(1-\text{cosec } A)} && \text{(rearrange)}\\ &=\frac{(\cot A-(1-\text{cosec } A))^2}{\cot^2A-(1-\text{cosec } A)^2} && \text{(multiply by conjugate)}\\ &=\frac{\text{cosec }^2A+\cot A\text{cosec } A-\text{cosec } A-\cot A}{\text{cosec } A-1} && \text{(expand)}\\ &=\text{cosec } A+\cot A && \text{(factorize and cancel)} \end{aligned} \]

vi.

\[ \frac{1+\sin A}{1-\sin A} \] \[ \begin{aligned} &=\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} && \text{(multiply by conjugate)}\\ &=\frac{1+\sin A}{\cos A} && \text{(use identity)}\\ &=\sec A+\tan A && \text{(split)} \end{aligned} \]

vii.

\[ \frac{\sin \theta-2\sin^3 \theta}{2\cos^3 \theta-\cos \theta} \] \[ \begin{aligned} &=\frac{\sin \theta(1-2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta-1)} && \text{(factorize)}\\ &=\frac{\sin \theta \cos 2\theta}{\cos \theta \cos 2\theta} && \text{(use identities)}\\ &=\frac{\sin \theta}{\cos \theta} && \text{(cancel)}\\ &=\tan \theta \end{aligned} \]

viii.

\[ (\sin A+\text{cosec } A)^2+(\cos A+\sec A)^2 \] \[ \begin{aligned} &=\sin^2A+\text{cosec }^2A+2\sin A\text{cosec } A +\cos^2A+\sec^2A+2\cos A\sec A && \text{(expand)}\\ &=\sin^2A+(1+\cot^2A)+2 +\cos^2A+(1+\tan^2A)+2 && \text{(use identities)}\\ &=1+6+\tan^2A+\cot^2A && \text{(simplify)}\\ &=7+\tan^2A+\cot^2A \end{aligned} \]

ix.

\[ (\text{cosec } A-\sin A)(\sec A-\cos A) \] \[ \begin{aligned} &=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) && \text{(convert)}\\ &=\frac{1-\sin^2A}{\sin A}\cdot\frac{1-\cos^2A}{\cos A} && \text{(simplify)}\\ &=\frac{\cos^2A}{\sin A}\cdot\frac{\sin^2A}{\cos A} && \text{(use identities)}\\ &=\sin A\cos A \end{aligned} \] RHS: \[ \frac{1}{\tan A+\cot A} =\frac{\sin A\cos A}{1} =\sin A\cos A \]

x.

\[ \frac{1+\tan^2A}{1+\cot^2A} =\frac{\sec^2A}{\text{cosec }^2A} =\frac{\sin^2A}{\cos^2A} =\tan^2A \] \[ \begin{aligned} \left(\frac{1-\tan A}{1-\cot A}\right)^2 &=\frac{\sec^2A-2\tan A}{\text{cosec }^2A-2\cot A} && \text{(expand)}\\ &=\frac{\frac{1-2\sin A\cos A}{\cos^2A}}{\frac{1-2\sin A\cos A}{\sin^2A}} && \text{(convert)}\\ &=\frac{\sin^2A}{\cos^2A} && \text{(cancel)}\\ &=\tan^2A \end{aligned} \]

Exam Insight

  • Each step carries marks — writing logic ensures full marks
  • Identity + algebra = scoring combination
  • Very frequent in CBSE and competitive exams
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Chapter Complete!

All 4 solutions for Introduction to Trigonometry covered.

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NCERT  ·  Class X  ·  Mathematics  ·  Chapter 8  ·  Exercise 8.3
Introduction to Trigonometry
Trigonometric Identities — Complete AI-Powered Learning Engine
Core Concept of Exercise 8.3

What is a Trigonometric Identity?

A trigonometric identity is an equation involving trigonometric ratios of an angle that holds true for all values of the angle for which both sides are defined. Unlike equations (true for specific angles), identities are universally true. Exercise 8.3 introduces the three fundamental Pythagorean Identities — each derived directly from the Pythagorean theorem applied to a right triangle.

📌 Geometric Foundation: In △ABC right-angled at B, with AB = adjacent, BC = opposite, AC = hypotenuse, the Pythagorean theorem gives AB² + BC² = AC². Dividing both sides by AC², AB²/AC² + BC²/AC² = 1, i.e. cos²A + sin²A = 1. The other two identities follow by dividing by AB² and BC² respectively.

The Three Pythagorean Identities — Derivation

  • Identity 1: Divide AB² + BC² = AC² by AC² → cos²A + sin²A = 1
  • Identity 2: Divide AB² + BC² = AC² by AB² → 1 + tan²A = sec²A
  • Identity 3: Divide AB² + BC² = AC² by BC² → cot²A + 1 = cosec²A

These three are also called the Pythagorean triplets of trigonometry.

Strategy for Proving Identities

  • Always work on one side only — either simplify LHS to RHS or RHS to LHS, never both simultaneously.
  • Convert everything to sin and cos first — the most reliable approach for complex expressions.
  • Look for opportunities to use a² − b² = (a+b)(a−b) to factorise trig expressions.
  • When you see 1 ± sin, 1 ± cos, sec² − tan², look for known identities immediately.
  • In fractions, try rationalising or finding a common denominator.
📊
Identity Quick-Reference Table
All three identities and their key derived forms
Root Identity Rearrangement 1 Rearrangement 2 Valid When
sin²A + cos²A = 1 sin²A = 1 − cos²A cos²A = 1 − sin²A All A
1 + tan²A = sec²A sec²A − tan²A = 1 tan²A = sec²A − 1 A ≠ 90°
1 + cot²A = cosec²A cosec²A − cot²A = 1 cot²A = cosec²A − 1 A ≠ 0°
Complete Formula Reference — Exercise 8.3
🔷
The Three Pythagorean Identities
Foundation of Exercise 8.3 — memorise all derived forms
sin²A + cos²A = 1
First Pythagorean Identity
Derived: sin A = √(1−cos²A)  |  cos A = √(1−sin²A)
1 + tan²A = sec²A
Second Pythagorean Identity
Derived: sec²A − tan²A = 1  |  tan²A = sec²A − 1
1 + cot²A = cosec²A
Third Pythagorean Identity
Derived: cosec²A − cot²A = 1  |  cot²A = cosec²A − 1
🔗
Factored Forms — Frequently Used in Proofs
Applying a² − b² = (a+b)(a−b) to trigonometric identities
sec²A − tan²A
= (secA + tanA)(secA − tanA)
Factored Form · Identity 2
= 1  (always)
cosec²A − cot²A
= (cosecA + cotA)(cosecA − cotA)
Factored Form · Identity 3
= 1  (always)
1 − sin²A = cos²A
1 − cos²A = sin²A
Substitution Forms · Identity 1
Most common in simplifications
(sinA + cosA)² = 1 + 2sinAcosA
Expansion with Identity 1
since sin²+cos²=1 inside expansion
(sinA − cosA)² = 1 − 2sinAcosA
Expansion with Identity 1
Useful in proving square-form identities
sin⁴A + cos⁴A
= 1 − 2sin²Acos²A
Fourth Power Form
Via (a²+b²) = (a+b)² − 2ab with a=sin², b=cos²
📋
Reciprocal & Quotient Identities
Essential supporting identities used alongside Pythagorean identities
tan A = sin A / cos A
Quotient Identity
cot A = cos A / sin A
Quotient Identity
sec A = 1 / cos A
Reciprocal Identity
cosec A = 1 / sin A
Reciprocal Identity
cot A = 1 / tan A
Reciprocal Identity
sin A · cosec A = cos A · sec A = tan A · cot A = 1
Product Identity
Smart Strategies & Mnemonics
✨
Expert Tips for Trigonometric Identity Problems
Proven techniques used in board exams and competitive papers
1
Always Work on One Side Only

A critical exam rule: pick the more complex side and simplify it to match the simpler side. Never add or subtract terms from both sides simultaneously — that's algebraic manipulation, not identity proving.

To prove: (1−sinA)/(1+sinA) = (secA − tanA)²
→ Work on RHS → expand → simplify → show = LHS ✓
2
Convert Everything to sin and cos First

When stuck, replace tan, cot, sec, cosec with their sin/cos equivalents. The resulting expression is usually easier to simplify using Identity 1 (sin²+cos²=1).

tan²A + 1 → sin²A/cos²A + 1 → (sin²A + cos²A)/cos²A → 1/cos²A = sec²A ✓
3
The "Difference of Squares" Weapon

Whenever you see sec²A − tan²A or cosec²A − cot²A, replace immediately with 1. If it's a product (secA+tanA)(secA−tanA), it also equals 1.

(secA + tanA)(secA − tanA) = sec²A − tan²A = 1
(cosecA + cotA)(cosecA − cotA) = cosec²A − cot²A = 1
4
Rationalisation Trick for Square Roots

Expressions with (1+sinA), (1−sinA), (1+cosA), (1−cosA) in numerator/denominator — multiply top and bottom by the conjugate to trigger the Pythagorean identity.

1/(1+sinA) × (1−sinA)/(1−sinA) = (1−sinA)/(1−sin²A) = (1−sinA)/cos²A
5
Add Fractions to Reveal Hidden Identities

Expressions like sinA/(1+cosA) + (1+cosA)/sinA — take LCM and add. The numerator often simplifies using sin²+cos²=1 to give 2/sinA = 2cosecA.

[sin²A + (1+cosA)²] / [sinA(1+cosA)]
= [sin²A + 1 + 2cosA + cos²A] / [sinA(1+cosA)]
= [1 + 1 + 2cosA] / [sinA(1+cosA)] = 2(1+cosA)/[sinA(1+cosA)] = 2/sinA = 2cosecA ✓
6
Memory Aid: "STC" — Square, Try, Convert

When facing any identity: Square both sides if needed (for expressions with √), Try direct substitution of identities, Convert to sin/cos if stuck.

7
The "Product to Sum" Trick for sinAcosA

If sinA + cosA = k, then squaring: sin²A + cos²A + 2sinAcosA = k² → 1 + 2sinAcosA = k² → sinAcosA = (k²−1)/2. Use this to find sinAcosA without knowing A.

If sinA + cosA = √2 → (√2)² = 1 + 2sinAcosA → sinAcosA = ½
8
Recognise the "1" Substitution Instantly

Whenever you see a lone "1" in a trig expression, ask: can I write it as sin²A+cos²A, or as sec²A−tan²A, or as cosec²A−cot²A? This substitution often unlocks the whole proof.

(secA + tanA − 1)/(tanA − secA + 1):
Replace the −1 with −(sec²A − tan²A) = −(secA+tanA)(secA−tanA) → factor → simplify
9
Spotting the LCM Approach

In problems like sinA/(cosecA−cotA) − sinA/(cosecA+cotA), immediately find LCM = cosec²A−cot²A = 1, making the calculation clean.

Common Mistakes & Misconceptions
⚠️
Errors That Cost Marks in Board Exams
Study these carefully — each is based on real student errors
Confusing the Three Identities with Each Other
✗ Wrong: sin²A + tan²A = 1 ✓ Right: sin²A + cos²A = 1   (only sin and cos pair for identity 1)

Only the three standard Pythagorean pairs hold: (sin,cos), (tan,sec), (cot,cosec). Never mix across these pairs.

Working on Both Sides Simultaneously
✗ Wrong: LHS = … → [add x to both sides] ← RHS is also modified ✓ Right: Take only LHS, simplify using identities, arrive at RHS without touching it

Proving an identity by cross-multiplication or rearrangement assumes what you're trying to prove (circular reasoning). Always work on one side only.

Square Root Sign Errors
✗ Wrong: sin A = ±√(1 − cos²A) in all contexts ✓ Right: For acute angle A in a right triangle, sin A = +√(1 − cos²A) (positive root)

In the NCERT Class X context, angles are acute (0° < A < 90°), so all trig ratios are positive. Always take the positive root unless stated otherwise.

Miswriting 1 + tan²A = sec²A as tan²A + 1 = cosec²A
✗ Wrong: 1 + tan²A = cosec²A ✓ Right: 1 + tan²A = sec²A  |  1 + cot²A = cosec²A

tan pairs with sec; cot pairs with cosec. This is the most common memory error. Remember: "tan → sec" (both have cos in denominator) and "cot → cosec" (both have sin in denominator).

Incorrect Squaring: (sinA + cosA)² = sin²A + cos²A
✗ Wrong: (sinA + cosA)² = sin²A + cos²A = 1 ✓ Right: (sinA + cosA)² = sin²A + 2sinAcosA + cos²A = 1 + 2sinAcosA

The middle term 2sinAcosA is omitted. Always expand (a+b)² = a² + 2ab + b² correctly before applying identities.

Treating cosecA as (sinA)⁻¹ incorrectly in expressions
✗ Wrong: sin A + cosec A = sin A + 1/sin A = (sin²A + 1)/sinA ← then incorrectly = 2/sinA ✓ Right: (sin²A + 1)/sinA — cannot simplify numerator to 2 unless sin²A = 1 (which requires specific A)

sin²A + 1 ≠ 2. This only equals 2 when sinA = 1 (A = 90°). The expression stays as (1 + sin²A)/sinA in general.

Forgetting the Domain Restriction
✗ Wrong: Using 1 + tan²A = sec²A when A = 90° ✓ Right: This identity is undefined at A = 90° (tan 90° and sec 90° are both undefined)

Identity 2 requires A ≠ 90°; Identity 3 requires A ≠ 0°. In board exam proofs, state the domain when relevant.

Not Simplifying Fully Before Claiming Proof
✗ Wrong: Stopping at sin²A/cos²A + 1 and saying "= sec²A hence proved" ✓ Right: Continue: (sin²A + cos²A)/cos²A = 1/cos²A = sec²A = RHS ✓ (each step shown)

Every step must be shown explicitly in NCERT board exams. Don't skip intermediate steps even if they seem obvious.

Concept-Building Practice Questions
Q 01
Prove: (1 − sin²A) · sec²A = 1
Prove
▼
LHS
Take LHS = (1 − sin²A) · sec²A
IDENTITY
Use: 1 − sin²A = cos²A (from Identity 1)
SUBSTITUTE
= cos²A · sec²A = cos²A · (1/cos²A)
SIMPLIFY
= 1 = RHS ✓
Proved: LHS = (1−sin²A)·sec²A = cos²A·sec²A = 1 = RHS
Q 02
Prove: tanA/(1 − cotA) + cotA/(1 − tanA) = 1 + tanA + cotA
Prove
▼
REWRITE
Let t = tanA, so cotA = 1/t. LHS becomes:
t/(1 − 1/t) + (1/t)/(1 − t) = t·t/(t−1) + (1/t)·1/(1−t)
SIMPLIFY
= t²/(t−1) − 1/[t(t−1)]
= [t³ − 1] / [t(t−1)]
FACTOR
Use a³ − 1 = (a−1)(a² + a + 1):
t³ − 1 = (t−1)(t² + t + 1)
= (t−1)(t² + t + 1) / [t(t−1)] = (t² + t + 1)/t
EXPAND
= t²/t + t/t + 1/t = t + 1 + 1/t = tanA + 1 + cotA = RHS ✓
Proved ✓   (Using algebraic manipulation and substitution)
Q 03
Prove: √[(1 + sinA)/(1 − sinA)] = secA + tanA
Prove
▼
LHS
LHS = √[(1 + sinA)/(1 − sinA)]
RATIONALISE
Multiply numerator and denominator inside √ by (1 + sinA):
= √[(1 + sinA)² / (1 − sin²A)]
IDENTITY
1 − sin²A = cos²A, so:
= √[(1 + sinA)² / cos²A] = (1 + sinA) / cosA   (taking positive root since acute angle)
SPLIT
= 1/cosA + sinA/cosA = secA + tanA = RHS ✓
Proved ✓
Q 04
Simplify: sin⁴A − cos⁴A + 1
Simplify
▼
FACTOR
sin⁴A − cos⁴A = (sin²A − cos²A)(sin²A + cos²A)
IDENTITY
sin²A + cos²A = 1, so:
sin⁴A − cos⁴A = (sin²A − cos²A) · 1 = sin²A − cos²A
SUBSTITUTE
sin²A − cos²A + 1 = sin²A − (1 − sin²A) + 1
= sin²A − 1 + sin²A + 1 = 2sin²A
Answer: 2sin²A
Q 05
If tanA + secA = m, show that sinA = (m² − 1)/(m² + 1)
Find ValuesProve
▼
GIVEN
tanA + secA = m  …(i)
IDENTITY
sec²A − tan²A = 1 → (secA + tanA)(secA − tanA) = 1
So secA − tanA = 1/m  …(ii)
ADD (i)+(ii)
2secA = m + 1/m = (m² + 1)/m → secA = (m² + 1)/(2m)
→ cosA = 2m/(m² + 1)
SUBTRACT (i)−(ii)
2tanA = m − 1/m = (m² − 1)/m → tanA = (m² − 1)/(2m)
FIND sinA
sinA = tanA × cosA = [(m² − 1)/(2m)] × [2m/(m² + 1)] = (m² − 1)/(m² + 1) ✓
Proved: sinA = (m² − 1)/(m² + 1)
Q 06
Prove: (cosecA − sinA)(secA − cosA) = 1/(tanA + cotA)
ProveSimplify
▼
LHS — EXPAND
(cosecA − sinA)(secA − cosA)
= (1/sinA − sinA)(1/cosA − cosA)
= [(1 − sin²A)/sinA] · [(1 − cos²A)/cosA]
= (cos²A/sinA) · (sin²A/cosA)
LHS — SIMPLIFY
= cos²A · sin²A / (sinA · cosA) = sinA · cosA
RHS — SIMPLIFY
1/(tanA + cotA) = 1/(sinA/cosA + cosA/sinA)
= 1/[(sin²A + cos²A)/(sinA·cosA)]
= sinA·cosA/1 = sinA·cosA
CONCLUDE
LHS = sinAcosA = RHS ✓
Proved ✓   Both sides equal sinA·cosA
Q 07
If sinA + cosA = √2 · cosA, find the value of cotA.
Find Values
▼
GIVEN
sinA + cosA = √2 · cosA
REARRANGE
sinA = √2 · cosA − cosA = cosA(√2 − 1)
FIND cotA
cotA = cosA/sinA = cosA / [cosA(√2 − 1)] = 1/(√2 − 1)
RATIONALISE
= (√2 + 1) / [(√2 − 1)(√2 + 1)] = (√2 + 1)/(2 − 1) = √2 + 1
Answer: cotA = √2 + 1
Q 08
Prove: sin⁶A + cos⁶A + 3sin²Acos²A = 1
Prove
▼
STRATEGY
Use a³ + b³ = (a + b)³ − 3ab(a + b) with a = sin²A, b = cos²A.
APPLY
sin⁶A + cos⁶A = (sin²A)³ + (cos²A)³
= (sin²A + cos²A)³ − 3sin²Acos²A(sin²A + cos²A)
= 1³ − 3sin²Acos²A(1) = 1 − 3sin²Acos²A
SUBSTITUTE
LHS = sin⁶A + cos⁶A + 3sin²Acos²A
= (1 − 3sin²Acos²A) + 3sin²Acos²A = 1 = RHS ✓
Proved ✓   (Using the sum of cubes identity)
Q 09
Simplify: (sec²A − 1)(cosec²A − 1)
SimplifyConceptual
▼
IDENTITY 1
sec²A − 1 = tan²A   (from Identity 2: 1 + tan²A = sec²A)
IDENTITY 2
cosec²A − 1 = cot²A   (from Identity 3: 1 + cot²A = cosec²A)
PRODUCT
(sec²A − 1)(cosec²A − 1) = tan²A · cot²A
SIMPLIFY
= tan²A · (1/tan²A) = 1
Answer: 1
Q 10
Prove: (sinA + cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A
ProveSimplify
▼
LHS — EXPAND
(sinA + cosecA)² = sin²A + 2sinA·cosecA + cosec²A = sin²A + 2 + cosec²A
(cosA + secA)² = cos²A + 2cosA·secA + sec²A = cos²A + 2 + sec²A
ADD
LHS = sin²A + cos²A + 4 + sec²A + cosec²A
= 1 + 4 + sec²A + cosec²A   (using Identity 1)
SUBSTITUTE
sec²A = 1 + tan²A   and   cosec²A = 1 + cot²A
= 5 + (1 + tan²A) + (1 + cot²A) = 5 + 2 + tan²A + cot²A
CONCLUDE
= 7 + tan²A + cot²A = RHS ✓
Proved ✓   LHS = 7 + tan²A + cot²A = RHS
Q 11
If cosA = 3/5, find all other trigonometric ratios of A.
Find Values
▼
GIVEN
cosA = 3/5
FIND sinA
sin²A = 1 − cos²A = 1 − 9/25 = 16/25 → sinA = 4/5
FIND tanA
tanA = sinA/cosA = (4/5)/(3/5) = 4/3
RECIPROCALS
cotA = 3/4  |  secA = 5/3  |  cosecA = 5/4
VERIFY
sin²A + cos²A = 16/25 + 9/25 = 25/25 = 1 ✓
1 + tan²A = 1 + 16/9 = 25/9 = (5/3)² = sec²A ✓
sinA=4/5, tanA=4/3, cotA=3/4, secA=5/3, cosecA=5/4
Q 12
Prove: (tanA + secA − 1)/(tanA − secA + 1) = (1 + sinA)/cosA
ProveConceptual
▼
STRATEGY
Replace −1 in numerator using sec²A − tan²A = 1, so −1 = tan²A − sec²A = (tanA−secA)(tanA+secA)
NUMERATOR
tanA + secA − 1 = tanA + secA + (tanA − secA)(tanA + secA)
= (tanA + secA)[1 + tanA − secA]
DENOMINATOR
tanA − secA + 1 = 1 + tanA − secA   (same factor as inside bracket)
CANCEL
LHS = (tanA + secA)(1 + tanA − secA) / (1 + tanA − secA) = tanA + secA
= sinA/cosA + 1/cosA = (1 + sinA)/cosA = RHS ✓
Proved ✓   LHS = tanA + secA = (1+sinA)/cosA = RHS
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Introduction To Trigonometry | Mathematics Class -10
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