JEE Main 2025 Jan 28 Shift 1, Mathematics Q10: Inverse Trigonometric Functions

Rewrite each ${\sin }^{-1}$ as ${\tan }^{-1}$ using ${\sin }^{-1}(p/h)={\tan }^{-1}(p/\sqrt{h^2-p^2})$:

${\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{3}{4}$, ${\sin }^{-1}\frac{5}{13}={\tan }^{-1}\frac{5}{12}$, ${\sin }^{-1}\frac{33}{65}={\tan }^{-1}\frac{33}{56}$.

Combine the first two with the ${\tan }^{-1}$ addition formula:

${\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{5}{12}={\tan }^{-1}\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}={\tan }^{-1}\frac{56/48}{33/48}={\tan }^{-1}\frac{56}{33}$.

So the argument becomes ${\tan }^{-1}\frac{56}{33}+{\tan }^{-1}\frac{33}{56}$. Since $\frac{56}{33}$ and $\frac{33}{56}$ are reciprocals of positive numbers, their ${\tan }^{-1}$ values sum to $\frac{\pi }{2}$.

$\therefore \cos \left(\frac{\pi }{2}\right)=0$.