Question

If ${\sin }^{-1}\frac{\alpha }{17}+{\cos }^{-1}\frac{4}{5}-{\tan }^{-1}\frac{77}{36}=0,0<\alpha <13$, then ${\sin }^{-1}(\sin \alpha )+{\cos }^{-1}(\cos \alpha )$ is equal to

• A. $16-5\pi$
• B. 16
• C. 0
• D. $\pi$

Answer 1

Answer: (D)

${\cos }^{-1}\frac{4}{5}={\tan }^{-1}\frac{3}{4}$
$\therefore {\sin }^{-1}\frac{\alpha }{17}={\tan }^{-1}\frac{77}{36}-{\tan }^{-1}\frac{3}{4}={\tan }^{-1}\left(\frac{\frac{77}{36}-\frac{3}{4}}{1+\frac{77}{36}\cdot \frac{3}{4}}\right)$
${\sin }^{-1}\frac{\alpha }{17}={\tan }^{-1}\frac{8}{15}={\sin }^{-1}\frac{8}{17}$
$\Rightarrow \frac{\alpha }{17}=\frac{8}{17}\Rightarrow \alpha =8$
$\therefore {\sin }^{-1}(\sin 8)+{\cos }^{-1}(\cos 8)$
$=3\pi -8+8-2\pi$
$=\pi$