Question

If ${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=a\pi$.
Find the value of $a$.

• A. $a=\frac{1}{4}$
• B. $a=\frac{1}{2}$
• C. $a=1$
• D. $a=2$

Answer 1

The correct option is C $a=1$

Given: ${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=a\pi$ ..... $\left(i\right)$

${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
$={\tan }^{-1}\left(\frac{12}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
$={\tan }^{-1}\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}.\frac{3}{4}}+{\tan }^{-1}\left(\frac{63}{16}\right)={\tan }^{-1}\frac{-63}{16}+{\tan }^{-1}\left(\frac{63}{16}\right)$
$=\pi -{\tan }^{-1}\left(\frac{63}{16}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=\pi$

$\therefore {\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)=\pi$

Comparing with $\left(i\right)$, we get

$a=1$.