2 Marks Questions

Question 12 Marks

Find the value of $\tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right)$.

Answer

Suppose $\cos ^{-1} \frac{2}{\sqrt{5}}=\theta$ then $\cos \theta=\frac{2}{\sqrt{5}}$$
\begin{aligned}
\therefore \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right) & =\tan \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \\
& =\sqrt{\frac{1-2 / 55}{1+2 / 55}}=\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}} \\
& =\sqrt{\frac{(\sqrt{5}-2)^2}{5-4}}=\sqrt{5}-2 \text {}
\end{aligned}
$

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Question 22 Marks

Find the value of $\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)$.

Answer

$
\begin{aligned}
\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right) & =\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right] \\
& =\sin ^{-1}\left[\cos \frac{3 \pi}{5}\right] \\
& =\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right] \\
& =\sin ^{-1}\left[\sin \left(\frac{-\pi}{10}\right)\right] \\
& =\frac{-\pi}{10} \text { }
\end{aligned}
$

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Question 32 Marks

Find the value of $\sec ^{-1}(-2)-\sin ^{-1}\left(\frac{1}{2}\right)$.

Answer

$\sec ^{-1}(-2)-\sin ^{-1}\left(\frac{1}{2}\right)$
We know that :
$
\begin{aligned}
\sin ^{-1}\left(\frac{1}{2}\right) & =\frac{\pi}{6} \\
\cos \frac{2 \pi}{3} & =-\frac{1}{2} \\
\therefore \quad \sec \frac{2 \pi}{3} & =-2 \quad \therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}
\end{aligned}
$
Putting the value
$\Rightarrow \quad \frac{2 \pi}{3}-\frac{\pi}{6}=\frac{4 \pi-\pi}{6}=\frac{3 \pi}{6}=\frac{\pi}{2}$

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Question 42 Marks

Prove that $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$

Answer

$\text{L.H.S.}=\tan ^{-1} \frac{2}{3}= A ($suppose$)$
or
$\tan A=\frac{2}{3}$
$\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}$
$=\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^2}=\frac{\frac{4}{3}}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}$
$\Rightarrow \quad \tan 2 A=\frac{4}{3} \times \frac{9}{5}=\frac{12}{5}$
$\Rightarrow \quad 2 A=\tan ^{-1}\left(\frac{12}{5}\right)$
$\Rightarrow \quad A =\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right)$
$\Rightarrow \quad \tan ^{-1}\left(\frac{2}{3}\right)=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \quad$
Hence proved.

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Question 52 Marks

If $\sin ^{-1} x=\frac{1}{2}$ then write the value of $\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\}$.

Answer

$
\begin{aligned}
\sin ^{-1}\left\{2 x \sqrt{1-x^2}\right\} & =\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{1-\sin ^2\left(\frac{1}{2}\right)}\right\} \\
& {\left[\because \text { given } \sin ^{-1}(x)=\frac{1}{2}\right] } \\
& =\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \sqrt{\cos ^2\left(\frac{1}{2}\right)}\right\} \\
& =\sin ^{-1}\left\{2 \sin \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}\right)\right\} \\
& =\sin ^{-1}\left\{\sin \left(2 \times \frac{1}{2}\right)\right\} \\
& =\sin ^{-1}(\sin (1)) \\
& =1 \text {}
\end{aligned}
$

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Question 62 Marks

If $\sin ^{-1}\left(\frac{3}{4}\right)+\sec ^{-1}\left(\frac{4}{3}\right)=x$ then find the value of $x$.

Answer

$\sin ^{-1}\left(\frac{3}{4}\right)+\sec ^{-1}\left(\frac{4}{3}\right)=x$
$\Rightarrow \sin ^{-1}\left(\frac{3}{4}\right)+\cos ^{-1}\left(\frac{3}{4}\right)=x, \quad\left[\because \sec ^{-1} x=\cos ^{-1} \frac{1}{x}\right]$
$\therefore x=\frac{\pi}{2},\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

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Question 72 Marks

Find the value of $\sin \left[2 \sin ^{-1}(0.6)\right]$

Answer

We know that $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$, if
$\frac{-1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$
$\therefore 2 \sin ^{-1}(0.6)=\sin ^{-1}(2 \times 0.6 \times \sqrt{1-0.36})$
$=\sin ^{-1}(0.96)$
$\Rightarrow \sin \left(2 \sin ^{-1}(0.6)\right)=\sin \left(\sin ^{-1}(0.96)\right)$
$=0.96 \text { }$

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