Inverse Trigonometric Functions – Summation Property – JEE Main 28 Jan 2025 Shift 1

Solution:

Let $E = \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{33}{65}$.

Convert terms to $\tan^{-1}$:
1) $\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$
2) $\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}$
3) $\sin^{-1}\frac{33}{65} = \tan^{-1}\frac{33}{56}$

First, add the first two terms:
$$ \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{5}{12} = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{5}{12}}{1 – \frac{3}{4} \cdot \frac{5}{12}} \right) $$
$$ = \tan^{-1}\left( \frac{\frac{9+5}{12}}{1 – \frac{15}{48}} \right) = \tan^{-1}\left( \frac{\frac{14}{12}}{\frac{33}{48}} \right) $$
$$ = \tan^{-1}\left( \frac{14}{12} \times \frac{48}{33} \right) = \tan^{-1}\left( \frac{14 \times 4}{33} \right) = \tan^{-1}\left( \frac{56}{33} \right) $$

Now the expression becomes:
$$ \cos\left( \tan^{-1}\frac{56}{33} + \sin^{-1}\frac{33}{65} \right) $$
We know $\sin^{-1}\frac{33}{65} = \cot^{-1}\frac{56}{33}$ (Since $\tan \theta = 33/56$, $\cot \theta = 56/33$).
So,
$$ \cos\left( \tan^{-1}\frac{56}{33} + \cot^{-1}\frac{56}{33} \right) $$
Using the property $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$:
$$ \cos\left( \frac{\pi}{2} \right) = 0 $$

Ans. (2)