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.