Question

Solve: ${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is A $\frac{\pi }{2}$

${\sin }^{-1}\left(\frac{4}{5}\right)+{\sin }^{-1}\left(\frac{5}{13}\right)+{\sin }^{-1}\left(\frac{16}{65}\right)=\frac{\pi }{2}$

${\sin }^{-1}A+{\sin }^{-1}B={\sin }^{-1}\left(A\sqrt{1-B^{.2}}+B\sqrt{1-B^2}\right)$

${\sin }^{-1}\left(\frac{4}{5}\sqrt{1-{\left(\frac{5}{12}\right)}^2}+\frac{5}{13}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right)$

${\sin }^{-1}\left(\frac{63}{65}\right)$

${\sin }^{-1}\left(\frac{63}{65}\right)+{\sin }^{-1}\frac{16}{65}$

Applying some formula

${\sin }^{-1}\left(\frac{63}{65}\sqrt{1-{\left(-\frac{16}{65}\right)}^2}+\frac{16}{65}\sqrt{1-{\left(\frac{63}{65}\right)}^2}\right)$

${\sin }^{-1}\left(1\right)$

${\sin }^{-1}\left(\sin \pi /2\right)$

$\pi /2$