Exercise 8.2 (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 8.2 Question 1.

Evaluate:
(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$
(ii) $2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}$
(iii) $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$
(iv) $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
(v) $\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}$

Answer.

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$
$
\begin{aligned}
& =\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2} \\
& =\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) } 2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ} \\
& =2(1)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2 \\
& =2+\frac{3}{4}-\frac{3}{4}=2
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& \text { (iii) } \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}} \\
& =\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}} \\
& =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2+2 \sqrt{3}}=\frac{\sqrt{3}}{\sqrt{2} \times 2(\sqrt{3}+1)} \\
& =\frac{\sqrt{3}}{\sqrt{2} \times 2(\sqrt{3}+1)} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\
& =\frac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{2} \times 2(3-1)} \\
& =\frac{\sqrt{3}(\sqrt{3}-1)}{4 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\
& =\frac{3 \sqrt{2}-\sqrt{6}}{8}
\end{aligned}\\
&\text { (iv) } \begin{aligned}
& \frac{\sin 30^{\circ}+\tan 45^{\circ}-\cos e c 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}} \\
= & \frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}
\end{aligned}
\end{aligned}$

$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}$

$
\begin{aligned}
& =\frac{27+16-24 \sqrt{3}}{27-16} \quad\left[\text { SSince }(a+b)(a-b)=a^2-b^2\right] \\
& =\frac{43-24 \sqrt{3}}{11}
\end{aligned}
$
(v) $\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}$
$
\begin{aligned}
& \frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \\
= & \frac{5 \times \frac{1}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{4}{4}} \\
= & \frac{15+64-12}{12}=\frac{67}{12}
\end{aligned}
$

Ex 8.2 Question 2.

Choose the correct option and justify:
(i) $\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=$
(A) $\sin 60^{\circ}$
(B) $\cos 60^{\circ}$
(C) $\tan 60^{\circ}$
(D) $\sin 30^{\circ}$

(ii) $\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=$
(A) $\tan 90^{\circ}$
(B) 1
(C) $\sin 45^{\circ}$
(D) 0
(iii) $\sin 2 A=2 \sin A$ is true when $\mathbf{A}=$
(A) $0^{\circ}$
(B) $30^{\circ}$
(C) $45^{\circ}$
(D) $60^{\circ}$
(iv) $\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=$
(A) $\cos 60^{\circ}$
(B) $\sin 60^{\circ}$
(C) $\tan 60^{\circ}$
(D) None of these

Answer.

(i) (A) $\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}$

$=\frac{2 \times 1 / \sqrt{3}}{1+(1 / \sqrt{3})^2}$

$
=\frac{2}{\sqrt{3}} \times \frac{3}{3+1}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}
$
(ii) (D) $\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-1}{1+1}=\frac{0}{2}=0$
(iii) (A) Since $A=0$, then
$
\begin{aligned}
& \sin 2 A=\sin 0^{\circ}=0 \text { and } 2 \sin A=2 \sin 0^{\circ} \\
& =2 \times 0=0 \\
& \therefore \sin 2 A=\sin A \text { when } \mathrm{A}=0
\end{aligned}
$
$
\begin{aligned}
& \text { (iv) (C) } \frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}} \\
& =\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}} \\
& =\frac{2}{\sqrt{3}} \times \frac{3}{3-1}=\sqrt{3}=\tan 60^{\circ}
\end{aligned}
$

Ex 8.2 Question 3.

If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}} ; 0^{\circ}B$, find $A$ and $B$.

Answer.
$\tan (A+B)=\sqrt{3}$
$
\begin{aligned}
& \Rightarrow \tan (A+B)=\tan 60^{\circ} \\
& \Rightarrow \mathrm{A}+\mathrm{B}=60^{\circ} \ldots \ldots \ldots \text { (i) }
\end{aligned}
$

Also, $\quad \tan (A-B)=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan (A-B)=\tan 30^{\circ}$
$\Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}$.
On adding eq. (i) and (ii), we get,
$
2 \mathrm{~A}=90^{\circ} \Rightarrow \mathrm{A}=45^{\circ}
$

On Subtracting eq. (i) and eq. (ii), we get
$
2 \mathrm{~B}=30^{\circ} \Rightarrow \mathrm{B}=15^{\circ}
$
Ex 8.2 Question 4.

State whether the following are true or false. Justify your answer.
(i) $\sin (A+B)=\sin A+\sin B$
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
(iv) $\sin \theta=\cos \theta$ for all values of $\theta$.
(v) $\cot A$ is not defined for $A=0^{\circ}$.

Answer.

(i) False, because, let $\mathrm{A}=60^{\circ}$ and $\mathrm{B}=30^{\circ}$
Then, $\sin (A+B)=\sin \left(60^{\circ}+30^{\circ}\right)=\sin 90^{\circ}=1$
And $\sin A+\sin B=\sin 60^{\circ}+\sin 30^{\circ}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}$
$
\therefore \sin (A+B) \neq \sin A+\sin B
$
(ii) True, because it is clear from the table below:

Therefore, it is clear, the value of $\sin \theta$ increases as $\theta$ increases.
(iii) False, because

It is clear, the value of $\cos \theta$ decreases as $\theta$ increases
(iv) False as it is only true for $\theta=45^{\circ}$.
$
\Rightarrow \sin 45^{\circ}=\frac{1}{\sqrt{2}}=\cos 45^{\circ}
$
(v) True, because $\tan 0^{\circ}=0$ and $\cot 0^{\circ}=\frac{1}{\tan 0^{\circ}}$
$
=\frac{1}{0} \text { i.e. undefined. }
$