Question

If ${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{a}$.
Find the vale of $a$.

• A. $a=+1$
• B. $a=-1$
• C. $a=+2$
• D. $a=-2$

Answer 1

Answer: (C)

$a=+2$

Given, ${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{a}$
$\Rightarrow {\sin }^{-1}(\frac{4}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}+\frac{5}{13}\sqrt{1-{\left(\frac{4}{5}\right)}^2})+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{a}$
$\Rightarrow {\sin }^{-1}\left(\frac{63}{65}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{a}$
$\Rightarrow {\sin }^{-1}(\frac{63}{65}\times \sqrt{1-{\left(\frac{16}{65}\right)}^2}+\frac{16}{65}\sqrt{1-{\left(\frac{63}{65}\right)}^2})=\frac{\pi }{a}$
$\Rightarrow {\sin }^{-1}\left(1\right)=\frac{\pi }{a}\Rightarrow a=2$