Exercise 8.3 (Revised) - Chapter 8 - Introduction To Trigonometry - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths: Chapter 8 - Introduction to Trigonometry Solutions

Ex 8.3 Question 1.

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

Answer.

For $\sin A$,
By using identity $\operatorname{cosec}^2 A-\cot ^2 A=1$
$
\begin{aligned}
& \Rightarrow \operatorname{cosec} A=1+\cot ^2 A \\
& \Rightarrow \frac{1}{\sin ^2 A}=1+\cot ^2 A \\
& \Rightarrow \sin ^2 A=\frac{1}{1+\cot ^2 A} \\
& \Rightarrow \sin A=\frac{1}{\sqrt{1+\cot ^2 A}}
\end{aligned}
$

For $\sec A$,
By using identity $\sec ^2 A-\tan ^2 A=1$
$
\begin{aligned}
& \Rightarrow \sec ^2 A=1+\tan ^2 A \\
& \Rightarrow \sec ^2 A=1+\frac{1}{\cot ^2 A}=\frac{\cot ^2 A+1}{\cot ^2 A} \\
& \Rightarrow \sec ^2 A=\frac{1+\cot ^2 A}{\cot ^2 A} \\
& \Rightarrow \sec A=\frac{\sqrt{1+\cot ^2 A}}{\cot A}
\end{aligned}
$

For $\tan A$,
$
\tan A=\frac{1}{\cot A}
$
Ex 8.3 Question 2.

Write the other trigonometric ratios of $\mathrm{A}$ in terms of $\sec A$

Answer.

For $\sin A=$
By using identity, $\sin ^2 A+\cos ^2 A=1$
$
\begin{aligned}
& \Rightarrow \sin ^2 A=1-\cos ^2 A \\
& \Rightarrow \sin ^2 A=1-\frac{1}{\sec ^2 A}=\frac{\sec ^2 A-1}{\sec ^2 A} \\
& \Rightarrow \sin A=\frac{\sqrt{\sec ^2 A-1}}{\sec A}
\end{aligned}
$

For $\cos A$,
$
\cos A=\frac{1}{\sec A}
$

For $\tan A$,
By using identity $\sec ^2 A-\tan ^2 A=1$
$
\begin{aligned}
& \Rightarrow \tan ^2 A=\sec ^2 A-1 \\
& \Rightarrow \tan A=\sqrt{\sec ^2 A-1}
\end{aligned}
$

For $\operatorname{cosec} A$,

$\operatorname{cosec} A=\frac{1}{\sin A}=\frac{1}{\frac{\sqrt{\sec ^2 A-1}}{\sec A}}$

$
\Rightarrow \operatorname{cosec} A=\frac{\sec A}{\sqrt{\sec ^2 A-1}}
$

For $\cot A$,
$
\begin{aligned}
& \cot A=\frac{1}{\tan A} \\
& \Rightarrow \cot A=\frac{1}{\sqrt{\sec ^2 A-1}}
\end{aligned}
$
Ex 8.3 Question 3.

Evaluate:
(i) $\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}$
(ii) $\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}$

Answer.

(i) $\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}$
$
\begin{aligned}
& =\frac{\sin ^2 63^{\circ}+\sin ^2\left(90^{\circ}-63^{\circ}\right)}{\cos ^2\left(90^{\circ}-73^{\circ}\right)+\cos ^2 73^{\circ}} \\
& =\frac{\sin ^2 63^{\circ}+\cos ^2 63^{\circ}}{\sin ^2 73^{\circ}+\cos ^2 73^{\circ}} \\
& {\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta, \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]} \\
& =\frac{1}{1}=1\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]
\end{aligned}
$
(ii) $\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}$

$\begin{aligned}
& =\sin 25^{\circ} \cdot \cos \left(90^{\circ}-25^{\circ}\right)+\cos 25^{\circ} \cdot \sin \left(90^{\circ}-25^{\circ}\right) \\
& =\sin 25^{\circ} \cdot \sin 25^{\circ}+\cos 25^{\circ} \cdot \cos 25^{\circ}
\end{aligned}$

$
\begin{aligned}
& {\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta, \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]} \\
& =\sin ^2 25^{\circ}+\cos ^2 25^{\circ}=1 \\
& {\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]}
\end{aligned}
$
Ex 8.3 Question 4.

Choose the correct option. Justify your choice:
(i) $9 \sec ^2 A-9 \tan ^2 A=$
(A) 1
(B) 9
(C) 8
(D) 0
(ii) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$
(A) 0
(B) 1
(C) 2
(D) none of these
(iii) $(\sec A+\tan A)(1-\sin A)=$
(A) $\sec A$
(B) $\sin A$
(C) $\operatorname{cosec} A$

$\text { (D) } \cos A$

(iv) $\frac{1+\tan ^2 A}{1+\cot ^2 A}=$
(A) $\sec ^2 A$
(B) -1
(C) $\cot ^2 A$
(D) none of these
$
\begin{aligned}
& \text { Answer. (i) (B) } 9 \sec ^2 A-9 \tan ^2 A \\
& =9\left(\sec ^2 A-\tan ^2 A\right) \\
& =9 \times 1=9 \quad\left[\text { Since } \sec ^2 \theta-\tan ^2 \theta=1\right]
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }(\text { C) }(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta) \\
& =\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\
& =\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right) \\
& =\frac{(\cos \theta+\sin \theta)^2-(1)^2}{\cos \theta \cdot \sin \theta} \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta+2 \cos \theta \sin \theta-1}{\cos \theta \cdot \sin \theta} \\
& =\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \cdot \sin \theta} \\
& {\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]}
\end{aligned}
$

$=\frac{2 \cos \theta \sin \theta}{\cos \theta \cdot \sin \theta}=2$

$\begin{aligned}
& \text { (iii)(D) }(\sec A+\tan A)(1-\sin A) \\
& =\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A) \\
& =\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A) \\
& =\frac{1-\sin ^2 A}{\cos A} \quad\left[\text { Since }(a+b)(a-b)=a^2-b^2\right] \\
& =\frac{\cos ^2 A}{\cos A} \\
& =\cos A\left[\because 1-\sin ^2 A=\cos ^2 A\right] \\
& \text { (iv)(D) } \frac{1+\tan ^2 A}{1+\cot ^2 A}=\frac{\sec ^2 A-\tan ^2 A+\tan ^2 A}{\operatorname{cosec}^2 A-\cot ^2 A+\cot ^2 A} \\
& =\frac{\sec ^2 A}{\operatorname{cosec} A}=\frac{\frac{1}{\cos ^2 A}}{\frac{1}{\sin ^2 A}} \\
& =\frac{\sin ^2 A}{\cos ^2 A}=\tan ^2 A \\
&
\end{aligned}$

Ex 8.3 Question 5.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(i) $(\operatorname{cosec} \theta-\cot \theta)^2=\frac{1-\cos \theta}{1+\cos \theta}$
(ii) $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$

(iii) $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
(iv) $\frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}$

(v) $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$, using the identity $\operatorname{cosec} e^2 A=1+\cot ^2 A$
(vi) $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$
(vii) $\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}=\tan \theta$
(viii) $(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2$
$=7+\tan ^2 A+\cot ^2 A$
(ix) $(\operatorname{cosec} A-\sin A)(\sec A-\cos A)$
$
=\frac{1}{\tan A+\cot A}
$
(x) $\left(\frac{1+\tan ^2 A}{1+\cot ^2 A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\tan ^2 A$

Answer.

(i) L.H.S. $(\operatorname{cosec} \theta-\cot \theta)^2$
$
\begin{aligned}
& =\operatorname{cosec}^2 \theta+\cot ^2 \theta-2 \operatorname{cosec} \theta \cot \theta \quad\left[\text { Since }(a-b)^2=a^2+b^2-2 a b\right] \\
& =\frac{1}{\sin ^2 \theta}+\frac{\cos ^2 \theta}{\sin ^2 \theta}-2 \times \frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta}
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& =\frac{1+\cos ^2 \theta}{\sin ^2 \theta}-\frac{2 \cos \theta}{\sin ^2 \theta} \\
& =\frac{1+\cos ^2 \theta-2 \cos \theta}{\sin ^2 \theta} \\
& =\frac{(1-\cos \theta)^2}{\sin ^2 \theta}\left[\because a^2+b^2-2 a b=(a-b)^2\right] \\
& =\frac{(1-\cos \theta)(1-\cos \theta)}{1-\cos ^2 \theta} \\
& =\frac{(1-\cos \theta)(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)} \\
& =\frac{1-\cos \theta}{1+\cos \theta}=\text { R.H.S. }
\end{aligned}\\
&\begin{aligned}
& \text { (ii) L.H.S. } \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A} \\
& =\frac{\cos ^2 \theta+1+\sin ^2 \theta+2 \sin A}{(1+\sin A) \cos A} \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta+1+2 \sin A}{(1+\sin A) \cos A} \\
& =\frac{1+1+2 \sin A}{(1+\sin A) \cos A}\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\
& =\frac{2+2 \sin A}{(1+\sin A) \cos A}=\frac{2(1+\sin A)}{(1+\sin A) \cos A}
\end{aligned}
\end{aligned}$

$=\frac{2}{\cos A}=2 \sec A=\text { R.H.S }$

$\begin{aligned}
& \text { (iii) L.H.S. } \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} \\
& =\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}} \\
& =\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}} \\
& =\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta} \\
& =\frac{\sin ^2 \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^2 \theta}{\sin \theta(\cos \theta-\sin \theta)} \\
& =\frac{\sin ^2 \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^2 \theta}{\sin \theta(\sin \theta-\cos \theta)} \\
& =\frac{\sin ^3 \theta-\cos ^3 \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)} \\
& =\frac{(\sin \theta-\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta(\sin \theta-\cos \theta)} \\
& {\left[\because a^3-b^3=(a-b)\left(a^2+b^2+a b\right)\right]} \\
& =\frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta}\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right] \\
& =\frac{1}{\sin \theta \cos \theta}+\frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\
&
\end{aligned}$

$\begin{aligned}
& =\frac{1}{\sin \theta \cos \theta}+1=1+\frac{1}{\sin \theta \cos \theta} \\
& =1+\sec \theta \operatorname{cosec} \theta
\end{aligned}$

$
\begin{aligned}
& \text { (iv) L.H.S. } \frac{1+\sec A}{\sec A}=\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}=\frac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}} \\
& =\frac{\cos A+1}{\cos A} \times \frac{\cos A}{1}=1+\cos A \\
& =1+\cos A \times \frac{1-\cos A}{1-\cos A} \\
& =\frac{1-\cos ^2 A}{1-\cos ^2 A} \quad\left[\text { Since }(a+b)(a-b)=a^2-b^2\right] \\
& =\frac{\sin ^2 A}{1-\cos A}=\text { R.H.S. }
\end{aligned}
$
(v) L.H.S. $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing all terms by $\sin A$,
$
\begin{aligned}
& =\frac{\cot A-1+\operatorname{cosec} A}{\cot A+1-\operatorname{cosec} A}=\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1} \\
& =\frac{(\cot A+\operatorname{cosec} A)-\left(\operatorname{cosec}^2 A-\cot ^2 A\right)}{(1+\cot A-\operatorname{cosec} A)} \quad\left[\text { Since } \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right] \\
& =\frac{(\cot A+\operatorname{cosec} A)+\left(\cot ^2 A-\operatorname{cosec}^2 A\right)}{(1+\cot A-\cos e c A)} \\
& =\frac{(\cot \mathrm{A}+\operatorname{cosec} \mathrm{A})+(\cot \mathrm{A}+\operatorname{cosec} \mathrm{A})(\cot \mathrm{A}-\operatorname{cosec} \mathrm{A})}{(1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A})}
\end{aligned}
$

$=\cot A+\operatorname{cosec} A=\text { R.H.S. }$

$
\begin{aligned}
& \text { (vi) L.H.S. } \sqrt{\frac{1+\sin A}{1-\sin A}} \\
& =\sqrt{\frac{1+\sin A}{1-\sin A}} \times \sqrt{\frac{1+\sin A}{1+\sin A}} \\
& =\sqrt{\frac{(1+\sin A)^2}{1-\sin ^2 A}}\left[\because(a+b)(a-b)=a^2-b^2\right] \\
& =\sqrt{\frac{(1+\sin A)^2}{\cos ^2 A}}\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right] \\
& =\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A} \\
& =\sec A+\tan A=\text { R.H.S. }
\end{aligned}
$
(vii) L.H.S. $\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}$
$
\begin{aligned}
& =\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-1\right)} \\
& =\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left[2\left(1-\sin ^2 \theta\right)-1\right]} \\
& {\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right]} \\
& =\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(2-2 \sin ^2 \theta-1\right)}
\end{aligned}
$

$=\frac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(1-2 \sin ^2 \theta\right)}=\frac{\sin \theta}{\cos \theta}$

$\begin{aligned}
&\begin{aligned}
& =\tan \theta=\text { R.H.S } \\
& \text { (viii) L.H.S. }(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2 \\
& =\left(\sin A+\frac{1}{\sin A}\right)^2+\left(\cos A+\frac{1}{\cos A}\right)^2 \\
& =\sin ^2 A+\frac{1}{\sin ^2 A}+2 \cdot \sin A \cdot \frac{1}{\sin A}+\cos ^2 A+\frac{1}{\cos ^2 A}+2 \cdot \cos A \cdot \frac{1}{\cos A} \\
& =2+2+\sin ^2 A+\cos ^2 A+\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A} \\
& =4+1+\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A} \\
& =5+\operatorname{cosec}^2 A+\sec ^2 A \\
& =5+1+\cot ^2 A+1+\tan ^2 A \\
& {\left[\because \operatorname{cosec}^2 \theta=1+\cot ^2 \theta, \sec ^2 \theta=1+\tan ^2 \theta\right]} \\
& =7+\tan ^2 A+\cot ^2 A \\
& =\text { R.H.S. }
\end{aligned}\\
&\begin{aligned}
& \text { (ix) L.H.S. }(\operatorname{cosec} A-\sin A)(\sec A-\cos A) \\
& =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \\
& =\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)
\end{aligned}
\end{aligned}$

$=\frac{\cos ^2 A}{\sin A} \times \frac{\sin ^2 A}{\cos A}=\sin A \cdot \cos A$

$
=\frac{\sin A \cdot \cos A}{\sin ^2 A+\cos ^2 A}\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]
$

Dividing all the terms by $\sin A \cdot \cos A$,
$
\begin{aligned}
& =\frac{\frac{\sin A \cdot \cos A}{\sin A \cdot \cos A}}{\frac{\sin ^2 A}{\sin A \cdot \cos A}+\frac{\cos ^2 A}{\sin A \cdot \cos A}} \\
& =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}} \\
& =\frac{1}{\tan A+\cot A}=\text { R.H.S. } \\
& \text { (x) L.H.S. }\left(\frac{1+\tan ^2 A}{1+\cot ^2 A}\right)=\frac{\sec ^2 A}{\operatorname{cosec} e^2 A} \\
& {\left[\because 1+\tan ^2 \theta=\sec ^2 \theta, 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta\right]} \\
& =\frac{1}{\cos ^2 A} \times \frac{\sin ^2 A}{1}=\tan ^2 A=\text { R.H.S. } \\
& \text { Now, Middle side }=\left(\frac{1-\tan A}{1-\cot A}\right)^2=\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^2 \\
& =\left(\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right)^2 \\
& =\left(\frac{1-\tan \mathrm{A}}{\frac{-(1-\tan \mathrm{A})}{\tan \mathrm{A}}}\right)^2=(-\tan A)^2 \\
& =\tan ^2 A=\text { R.H.S. } \\
&
\end{aligned}
$