Solution of Inverse Trigonometric Functions – JEE Main 29 Jan 2025 Shift 1

Solution:

$\cos^{-1}x – \sin^{-1}x = \pi + \sin^{-1}(2x+1)$

Using $\cos^{-1}x – \sin^{-1}x = \cos^{-1}x – (\frac{\pi}{2} – \cos^{-1}x) = 2\cos^{-1}x – \frac{\pi}{2}$:

$2\cos^{-1}x – \frac{\pi}{2} = \pi + \sin^{-1}(2x+1)$

$2\cos^{-1}x = \frac{3\pi}{2} + \sin^{-1}(2x+1)$

$\cos(2\cos^{-1}x) = \cos\left(\frac{3\pi}{2} + \sin^{-1}(2x+1)\right)$

$2x^2 – 1 = \sin(\sin^{-1}(2x+1))$

$2x^2 – 1 = 2x + 1$

$2x^2 – 2x – 2 = 0 \Rightarrow x^2 – x – 1 = 0$

$x = \frac{1 \pm \sqrt{5}}{2}$

Domain check for $\sin^{-1}(2x+1)$:

$-1 \le 2x+1 \le 1 \Rightarrow -1 \le x \le 0$

$\therefore x = \frac{1-\sqrt{5}}{2}$ (since $\frac{1+\sqrt{5}}{2} > 1$)

Now, for $(2x-1)^2$:

$2x = 1 – \sqrt{5} \Rightarrow 2x – 1 = -\sqrt{5}$

$(2x-1)^2 = (-\sqrt{5})^2 = 5$

Ans. 5