Question

If ${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{a}{b}$.
Find the value of $a+b$.

• A. $9$
• B. $32$
• C. $98$
• D. $121$

Answer 1

Answer: (C)

$98$

Given: ${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{a}{b}$ ....... $\left(i\right)$

Now, ${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}$

$={\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}$
$={\cos }^{-1}(\frac{4}{5}.\frac{12}{13}-\frac{3}{5}.\frac{5}{13})$

$={\cos }^{-1}\frac{33}{65}$

${\sin }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

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

$a=33$ and $b=65$.

Hence, $a+b=98$.