Question

Statement I $\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{63}{16}$
Statement II The simplified form of $\cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right)$, where $x \in\left[-\frac{\pi}{2}, \frac{\pi}{4}\right]$ is $\tan ^{-1}\left(\frac{4}{3}\right)$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (A)

I. Let $\alpha=\sin ^{-1} \frac{5}{13}$ and $\beta=\cos ^{-1} \frac{3}{5}$
$ \Rightarrow \sin \alpha=\frac{5}{13} \text { and } \cos \beta=\frac{3}{5} $
$ \Rightarrow \cos \alpha=\sqrt{1-\frac{25}{169}} \text { and } \sin \beta=\sqrt{1-\frac{9}{25}} $
$ \Rightarrow \cos \alpha=\frac{12}{13} \text { and } \sin \beta=\frac{4}{5} $
$ \Rightarrow \tan \alpha=\frac{5}{12} \text { and } \tan \beta=\frac{4}{3}\left(\because \tan x=\frac{\sin x}{\cos x}\right)$
$ \Rightarrow \alpha=\tan ^{-1} \frac{5}{12} \text { and } \beta=\tan ^{-1} \frac{4}{3} $
Now, $ \sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$
$ =\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3}$
$ =\tan ^{-1}\left(\frac{15+48}{36-20}\right) $
$ =\tan ^{-1}\left(\frac{63}{16}\right)$
$\therefore$ Statement $I$ is true.
II. Let $\cos \alpha=\frac{3}{5}$
$\Rightarrow \sin \alpha=\sqrt{1-\frac{9}{25}}=\frac{4}{5} $
$\Rightarrow \tan \alpha=\frac{4}{3}$
Now, $ \cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right) $
$=\cos ^{-1}[\cos \alpha \cos x+\sin \alpha \sin x] $
$=\cos ^{-1}[\cos (\alpha-x)]=\alpha-x=\tan ^{-1} \frac{4}{3}-x$
$\therefore$ Statement II is false.