INTRODUCTION TO TRIGONOMETRY-Exercise 8.3
Maths - Exercise
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
sin A in terms of cot A
$$\begin{aligned}\cot A&=\dfrac{\cos A}{\sin A}\\\\ \sin A&=\dfrac{\cos A}{\cot A}\\\\ \color{magenta}\cos A&\color{magenta}=\sqrt{1-\sin^2 A}\\\\ &=\dfrac{\sqrt{1-\sin ^{2}A}}{\cot A}\\ \\ \sin ^{2}A&=\dfrac{1-\sin ^{2}A}{\cot ^{2}A}\\\\ \cot ^{2}A \sin ^{2}A&=1-\sin ^{2}A\\\\ \cot ^{2}A \sin ^{2}A+\sin ^{2}A&=1\\\\ \sin ^{2}A \left( \cot ^{2}A+1\right) &=1\\\\ \sin ^{2}A&=\dfrac{1}{1+\cot ^{2}A}\\\\ \sin A&=\sqrt{\dfrac{1}{1+\cot ^{2}A}}\\\\ &=\dfrac{1}{\sqrt{1+\cot ^{2}A}}\end{aligned}$$sec A in terms of cot A
$$\begin{aligned}\cot A&=\dfrac{\cos A}{\sin A}\\\\ \color{magenta}\sin A&\color{magenta}=\sqrt{1-\cos^2 A}\\\\ \cot A&=\dfrac{\cos A}{\sqrt{1-\cos 2}}\\\\ \dfrac{\sqrt{1-\cos ^{2}A}}{\cos A}&=\dfrac{1}{\cot A}\\\\ \left( \dfrac{\sqrt{1-\cos ^{2}A}}{\cos A}\right) ^{2}&=\left( \dfrac{1}{\cot A}\right) ^{2}\\\\ \dfrac{1-\cos ^{2}A}{\cos ^{2}A}&=\dfrac{1}{\cot ^{2}A}\\\\ \dfrac{1}{\cos ^{2}A}-\dfrac{\cos ^{2}A}{\cos ^{2}A}&=\dfrac{1}{\cot ^{2}A}\\\\ \dfrac{1}{\cos ^{2}A}&=\dfrac{1}{\cot ^{2}A}-1\\\\ \sec ^{2}A&=\dfrac{1}{\cot ^{2}A}-1\\\\ \sec A&=\sqrt{\dfrac{1-\cot ^{2}A}{\cot ^{2}A}}\\\\ \sec A&=\dfrac{\sqrt{1-\cot ^{2}A}}{\cot A}\end{aligned}$$tan A in terms of cot A
$$\tan A=\dfrac{1}{\cot A}$$2. Write all the other trigonometric ratios of A in terms of sec A.
Solution:
Trigonometrical Ratio in terms of sec A
sin A and cosec A in terms of sec A
$$\begin{aligned}\sin A&=\sqrt{1-\cos ^{2}A}\\\\ \sin ^{2}A&=1-\cos ^{2}A\\\\ \dfrac{\sin ^{2}A}{\cos ^{2}A}&=\dfrac{1-\cos ^{2}A}{\cos ^{2}A}\\\\ \dfrac{\sin ^{2}A}{\cos ^{2}A}&=\dfrac{1}{\cos ^{2}A}-1\\\\ \sin ^{2}A \sec ^{2}A&=\sec ^{2}A-1\\\\ \sin ^{2}A&=\dfrac{\sec ^{2}A-1}{\sec ^{2}A}\\\\ \sin A&=\dfrac{\sqrt{\sec ^{2}A-1}}{\sec A}\\\\ \text{cosec } A&=\dfrac{1}{\sin A}\\\\ \text{cosec } A&=\dfrac{\sec A}{\sqrt{\sec ^{2}A-1}}\end{aligned}$$cos A in terms of secA
$$\begin{aligned}\cos A&=\dfrac{1}{\sec A}\\ \sec A&=\dfrac{1}{\cos A}\end{aligned}$$tan A and cot A in terms of sec A
$$\begin{aligned}\tan A&=\dfrac{\sin A}{\cos A}\\\\ \sin A&=\dfrac{\sqrt{\sec ^{2}A-1}}{\sec A}\\\\ \cos A&=\dfrac{1}{\sec A}\\\\ \tan A&=\dfrac{\sqrt{\sec ^{2}A-1}}{\sec A \cos A}\\\\ \tan A&=\dfrac{\sqrt{\sec ^{2}A-1}}{\sec A \dfrac{1}{\sec A}}\\\\ \tan A&=\sqrt{\sec ^{2}A-1}\\\\ \cot A&=\dfrac{1}{\tan A}\\\\ \cot A&=\dfrac{1}{\sqrt{\sec ^{2}A-1}}\end{aligned}$$3. Choose the correct option. Justify your choice.
Solution:
Choose the correct options
i. $$\begin{aligned} &9\sec ^{2}A-9 \tan ^{2}A\\ &=9 \left( \sec ^{2}A-\tan ^{2}A\right) \\ &=9 \left[ \sec ^{2}A-\left( \sec ^{2}A-1\right) \right] \\ &=9 \left[ \sec ^{2}A-\sec ^{2}A+1\right] \\ &=9 \left[ 1\right] \\ &=9\end{aligned}$$(B) 9 is correct option
ii. $$\scriptsize\begin{aligned} &\left( 1+\tan \theta +\sec\theta \right) \left( 1+\cot \theta -\text{cosec }\theta \right) \\\\ &=\left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta }\right) \left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta }\right) \\\\ &=\left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta }\right) \left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta }\right) \\\\ &=\dfrac{1}{\sin \theta \cos \theta } \left[ \left( \cos \theta +\sin \theta +1\right) \left( \sin \theta + \cos \theta -1\right) \right] \end{aligned}$$Let \(\sin \theta + \cos \theta =a, \text{ and } 1=b\)
using Formula: \((a+b)(a-b)=a^2-b^2\)
(C) 2 is correct option
iii. $$\begin{aligned} &\left( \sec A+\tan A\right) \left( 1-\sin A\right) \\\\ &=\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right) \left( 1-\sin A\right) \\\\ &=\dfrac{1}{\cos A} \left[ \left( 1+\sin A\right) \left( 1-\sin A\right) \right] \\\\ &=\dfrac{1}{\cos A} \left[ 1-\sin ^{2}A\right] \\\\ &=\dfrac{1}{\cos A} \cos ^{2}A\\\\ &=\cos A\end{aligned}$$(D) is correct option
iv. $$\begin{aligned}&\dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}\\\\ &=\dfrac{\sec ^{2}A}{\text{cosec}^{2}\ A}\\\\ &=\dfrac{1}{\cos ^{2}A \dfrac{1}{\sin ^{2}A}}\\\\ &=\dfrac{\sin ^{2}A}{\cos ^{2}A}\\\\ &=\tan ^{2}A\end{aligned}$$(D) is correct option
4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
Solution:
i. \[\left( \text{cosec }\theta -\cot \theta \right) ^{2}=\dfrac{1-\cos \theta }{1+\cos \theta }\]LHS
$$\begin{aligned}&\left( \text{cosec } \theta -\cot \theta \right) ^{2}\\\\ &\left[ \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta }\right] ^{2}\\\\ &=\left[ \dfrac{1-\cos \theta }{\sin \theta }\right] ^{2}\\\\ &=\dfrac{\left( 1-\cos \theta \right) ^{2}}{\sin ^{2}\theta }\\\\ &=\dfrac{\left( 1-\cos \theta \right) ^{2}}{\left( 1-\cos ^{2}\theta \right) }\\\\ \color{cyan}a^{2}-b^{2}&\color{cyan}=\left( a+b\right) \left( a-b\right) \\\\ &=\dfrac{\left( 1-\cos \theta \right) ^{2}}{\left( 1+\cos \theta \right) \left( 1-\cos \theta \right) }\\\\ &=\dfrac{1-\cos \theta }{1+\cos \theta }=RHS\\\\ LHS&=RHS\end{aligned}$$Thus Proved.
ii. \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]
LHS
$$\begin{aligned} &=\dfrac{\cos ^{2}A+\left( 1+\sin A\right) ^{2}}{\left( 1+\sin A\right) \cos A}\\\\ &=\dfrac{\cos ^{2}A+1+\sin ^{2}A+2 \sin A}{\left( 1+\sin A\right) \left( \cos A\right) }\\\\ &=\dfrac{\cos ^{2}A+\sin ^{2}A+1+2 \sin A}{\left( 1+\sin A\right) \left( \cos A\right) }\\\\ &\color{cyan}\cos ^{2}A+\sin ^{2}A=1\\\\ &=\dfrac{1+1+2 \sin A}{\left( 1+\sin A\right) \cos A}\\\\ &=\dfrac{2+2 \sin A}{\cos A \left( 1+\sin A\right) }\\\\ &=\dfrac{2 \left( 1+\sin A\right) }{\cos A \left( 1+\sin A\right) }\\\\ &=\dfrac{2}{\cos A}\\\\ &=2\sec A=RHS\\\\ LHS=RHS\end{aligned}$$Thus Proved
iii. \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \text{ cosec } \theta \]
LHS
$$\scriptsize\begin{aligned} &=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\\\\ &=\dfrac{\sin \theta }{\cos \theta \left[ \dfrac{\sin \theta -\cos \theta }{\sin \theta }\right] }+\dfrac{\cos \theta }{\sin \theta \left[ \dfrac{\cos \theta -\sin \theta }{\cos \theta }\right] }\\\\ &=\dfrac{\sin ^{2}\theta }{\cos \theta \left( \sin \theta -\cos \theta \right) }+\dfrac{\cos ^{2}\theta }{\sin \theta \left( \cos \theta -\sin \theta \right) }\\\\ &=\dfrac{\sin ^{3}\theta \left( \cos \theta -\sin \theta \right) +\cos ^{3}\theta \left( \sin \theta -\cos \theta \right) }{\cos \theta \left( \sin \theta -\cos \theta \right) \sin \theta \left( \cos \theta -\sin \theta \right) }\\\\ &=\dfrac{\sin ^{3}\theta \left( \cos \theta -\sin \theta \right) -\cos ^{3}\theta \left( \cos \theta -\sin \theta \right) }{\sin \theta \cos \theta \left( \sin \theta -\cos \theta \right) \left( \cos \theta -\sin \theta \right) }\\\\ &=\dfrac{\left( \cos \theta -\sin \theta \right) \left( \sin ^{3}\theta -\cos ^{3}\theta \right) }{\left( \cos \theta -\sin \theta \right) \left( \sin \theta \cos \theta \right) \left( \sin \theta -\cos \theta \right)}\\\\ &=\dfrac{\sin ^{3}\theta -\cos ^{3}\theta }{\sin \theta \cos \theta \left( \sin \theta -\cos \theta \right) }\end{aligned}$$Using Formula: \(\color{cyan}\boxed{a^3-b^3=(a-b)(a^2+ab+b^2)}\)
$$\scriptsize\begin{aligned}&=\dfrac{\left( \sin \theta -\cos \theta \right) \left( \sin ^{2}\theta +\cos ^{2}\theta +\sin \theta \cos \theta \right) }{\left( \sin \theta -\cos \theta \right) \left( \sin \theta \cos \theta \right) }\\\\ &=\dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\\\\ &=\dfrac{1 }{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta}\\\\ &=\dfrac{1}{\sin \theta \cos \theta }+1\\\\ &1+\sec \theta \text{ cosec }\theta =RHS\\\\ LHS&=RHS\end{aligned}$$Thus proved.
iv \[\dfrac{1+\sec A}{\sec A}=\dfrac{\sin ^{2}A}{1-\cos A}\\\]
LHS
$$\begin{align} &\dfrac{1+\sec A}{\sec A}\\\\ &=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\\\\ &=\dfrac{\cos A+1}{\cos A \dfrac{1}{\cos A}}\\\\ &=\cos A+1\tag{A}\end{align}$$RHS
$$\begin{align}&\dfrac{\sin ^{2}A}{1-\cos A}\\\\ &=\dfrac{1-\cos ^{2}A}{1-\cos A}\\\\ &=\dfrac{\left( 1-\cos A\right) \left( 1+\cos A\right) }{1-\cos A}\\\\ &=\cos A+1\tag{B}\\\\ A&=B\end{align}$$Thus proved
v. $$\begin{aligned} \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\end{aligned}$$
Divide numerator and denominator by sin A
$$\begin{aligned}&\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\sin A}+\dfrac{1}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\sin A}-\dfrac{1}{\sin A}}\\\\ &=\dfrac{\cot A-1+\text{ cosec }A}{\cot A+1-\text{ cosec }A}\\\\ &=\dfrac{\cot A-\left( 1-\text{ cosec }A\right) }{\cot A+\left( 1-\text{ cosec }A\right) }\end{aligned}$$Multiply and divide by cot A-(1-cosec A)
$$\scriptsize\begin{aligned}&\dfrac{\cot A-\left( 1-\text{ cosec }A\right) }{\cot A+\left( 1-\text{ cosec }A\right) }\times \dfrac{\cot A-\left( 1-\text{ cosec }A\right) }{\cot A-\left( 1-\text{ cosec }A\right) }\\\\ &=\dfrac{\left[ \cot A-\left( 1-\text{ cosec }A\right) \right] ^{2}}{\cot ^{2}A-\left( 1-\text{ cosec } A\right) ^{2}}\\\\ &=\dfrac{\cot ^{2}A-2 \cot A \left( 1-\cos aA\right) +\left( 1-\text{cosec }A\right) ^{2}}{\cot ^{2}A-\left[ 1+\text{ cosec } ^{2}A-2 \text{ cosec }A\right] }\\\\ &=\dfrac{\cot ^{2}A-2 \cot A+2 \cot A \text{ cosec }A+1+\text{ cosec }^{2}A-2\text{ cosec }A}{\cot ^{2}A-1-\text{ cosec}^{2}\ A+2 \text{ cosec }A}\\\\ &\color{cyan}\boxed{\text{cosec }^{2}A-1=\cot ^{2}A}\\\\ &=\dfrac{\text{cosec }^{2}A-1-2 \cot A+2 \cot A \text{ cosec }A+1+\text{ cosec }^{2}A-2 \text{ cosec }A}{\text{cosec}^{2}\ A-1-1-\text{ cosec }^{2}A+2 \text{ cosec }A}\\\\ &=\dfrac{2\text{ cosec } ^{2}A-2 \cot A+2 \cot A \text{ cosec } A-2\text{ cosec } A}{2 \text{ cosec }A-2}\end{aligned}$$ $$\scriptsize\begin{aligned}&=\dfrac{\text{cosec } ^{2}A+\cot A \text{ cosec } A-\text{ cosec } A-\cot A}{\text{ cosec } -1}\\\\ &=\dfrac{\text{cosec } A\left( \text{ cosec } A+\cot A\right) -1 \left( \text{ cosec } A+\cot A\right) }{\text{ cosec } -1}\\\\ &=\dfrac{\left( \text{cosec } A+\cot A\right) \left( \text{cosec } -1\right) }{\left( \text{cosec } -1\right) }\\\\ &=\text{cosec } A+\cot A\end{aligned}$$vi. \[\dfrac{1+\sin A}{1-\sin A}=\sec A+\tan A\]
LHS
$$\begin{aligned} &\sqrt{\dfrac{1+\sin A}{1-\sin A} \times\dfrac{1+\sin A}{1+\sin A}}\\\\ &=\sqrt{\dfrac{\left( 1+\sin A\right) ^{2}}{1-\sin ^{2}A}}\\\\ &=\sqrt{\dfrac{\left( 1+\sin A\right) ^{2}}{\cos ^{2}A}}\\\\ &=\dfrac{1+\sin A}{\cos A}\\\\ &=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\\\\ &=\sec A+\tan A=RHS\\\\ LHS=RHS\end{aligned}$$Thus Proved
vii. \[\dfrac{\sin \theta -2 \sin ^{3}\theta }{2 \cos ^{3}\theta -\cos \theta }=\tan \theta\]
LHS
$$\begin{aligned}&\dfrac{\sin \theta \left( 1-2 \sin ^{2}\theta \right) }{\cos \theta \left( 2 \cos ^{2}\theta -1\right) }\\\\ &\color{cyan}\boxed{1-2 \sin ^{2}\theta =\cos 2\theta} \\ &\color{cyan}\boxed{2 \cos ^{2}\theta -1=\cos 2\theta} \\\\ &=\dfrac{\sin \theta \left( \cos 2\theta \right) }{\cos \theta \left( \cos 2\theta \right) }\\\\ &=\dfrac{\sin \theta }{\cos \theta }\\\\ &=\tan \theta =RHS\\\\ LHS&=RHS\end{aligned}$$Thus Proved
viii. \[\scriptsize\left( \sin A+\text{ cosec }A\right) ^{2}+\left( \cos A+\sec A\right) ^{2}=7+\tan ^{2}A+\cot ^{2}A\]
LHS
$$\scriptsize\begin{aligned} &\left( \sin A+\text{ cosec }A\right) ^{2}+\left( \cos A+\sec A\right) ^{2}\\\\ &=\left( \sin ^{2}A+\text{ cosec }^{2}A+2 \sin A \text{ cosec }A\right) \\ &\quad+\left( \cos ^{2}A+\sec^{2}A+2 \cos A \sec A\right) \\\\ &=\left( \sin ^{2}A+1+\cot ^{2}A+2 \times\dfrac{\sin A}{\sin A}\right) \\ &\quad+\left( \cos ^{2}A+1+\tan ^{2}A+2\times \dfrac{\cos A}{\sin A}\right) \\\\ &=\left( \sin ^{2}A+1+\cot ^{2}A+2\right) \\ &\quad+\left( \cos ^{2}A+1+\tan ^{2}A+2\right) \\\\ &=\sin ^{2}A+1+\cot ^{2}A+2\\ &\quad+\cos ^{2}A+1+\tan ^{2}A+2\\\\ &=\sin ^{2}A+\cos ^{2}A+6\\ &\quad+\tan ^{2}A+\cot ^{2}A\\\\ &=1+6+\tan ^{2}A+\cot ^{2}A\\\\ &=7+\tan ^{2}A+\cot ^{2}A=RHS\\\\ LHS=RHS\end{aligned}$$Hence Proved
ix. \[\scriptsize\left( \text{cosec } A-\sin A\right) \left( \sec A-\cos A\right) =\dfrac{1}{\tan A+\cot A}\]
LHS
$$\begin{aligned} &\left( \text{cosec } A-\sin A\right) \left( \sec A-\cos A\right) \\\\ &=\left( \dfrac{1}{\sin A}-\sin A\right) \left( \dfrac{1}{\cos A}-\cos A\right) \\\\ &=\left( \dfrac{1-\sin ^{2}A}{\sin A}\right) \left( \dfrac{1-\cos ^{2}A}{\cos A}\right) \\\\ &=1-\sin ^{2}A=\cos ^{2}A\\\\ &=1-\cos ^{2}A=\sin ^{2}A\\\\ &=\dfrac{\cos ^{2}A}{\sin A} \dfrac{\sin ^{2}A}{\cos A}\\\\ &=\sin A \cos A\end{aligned}$$RHS
$$\begin{aligned} &\dfrac{1}{\tan A+\cot A}\\\\ &=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin }}\\\\ &=\dfrac{1}{\dfrac{\sin ^{2}A+\cos ^{2}A}{\cos A \sin A}}\\\\ &\color{cyan}\sin ^{2}A+\cos ^{2}A=1\\\\ &=\dfrac{\cos A \sin A}{1}\\\\ &=\sin A \cos A=LHS\\\\ LHS&=RHS\end{aligned}$$Hence Proved
x. \[\dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}=\left[ \dfrac{1-\tan A}{1-\cot A}\right] ^{2}=\tan ^{2}A\]
LHS
$$\begin{aligned} &\dfrac{1+\tan ^{2}A}{1+\cot ^{2}A}\\\\ &=\dfrac{\sec ^{2}A}{\text{ cosec } ^{2}A}\\\\ &=\dfrac{\sin ^{2}A}{\cos ^{2}A}\\\\ &=\tan ^{2}A\end{aligned}$$Hence Proved
LHS
\[\left[ \dfrac{1-\tan A}{1-\cot A}\right] ^{2}\] $$\begin{aligned} &\qquad\dfrac{\left( 1-\tan \right) ^{2}}{\left( 1-\cot A\right) ^{2}}\\\\ & \dfrac{1+\tan ^{2}A-2 \tan A}{1+\cot ^{2}A-2 \cot A}\\\\ & \dfrac{\sec ^{2}A-2 \tan A}{\text{ cosec }^{2}A-2 \cot A}\end{aligned}$$Simplify Numerator first
$$\begin{aligned}&\sec ^{2}A-2 \tan A\\\\ &=\dfrac{1}{\cos ^{2}A}-\dfrac{2 \sin A}{\cos A}\\\\ &=\dfrac{1-2 \sin A \cos A}{\cos ^{2}A}\end{aligned}$$Then simplify the Denominator
$$\begin{aligned}&\text{cosec }^{2}A-2 \cot A\\\\ &=\dfrac{1}{\sin ^{2}A}-\dfrac{2 \cos A}{\sin A}\\\\ &=\dfrac{1-2 \sin A \cos A}{\sin ^{2}A}\end{aligned}$$Now put together
$$\begin{aligned}&=\dfrac{\dfrac{1-2 \sin A \cos A}{\cos ^{2}A}}{\dfrac{1-2 \sin A \cos A}{\sin ^{2}A}}\\\\ &=\dfrac{\sin ^{2}A \left( 1-2 \sin A \cos A\right) }{\cos ^{2}A \left( 1-2 \sin A \cos A\right) }\\\\ &=\dfrac{\sin ^{2}A}{\cos ^{2}A}\\\\ &=\tan ^{2}A=RHS\end{aligned}$$ Thus Proved.
