JEE Main 2026 — 4 April, Morning Shift — Question 39

$y=\tan ^{-1}\left(\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4} \tan x}\right)+2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)$ For $2 \tan ^{-1}\left(\frac{\mathrm{x}}{1+\sqrt{1-\mathrm{x}^{2}}}\right)$ put $\mathrm{x}=\sin \theta$ $\therefore 2 \tan ^{-1}\left(\frac{\sin \theta}{1+\cos \theta}\right)=\theta=\sin ^{-1} \mathrm{x}$ Hence $y=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}(\tan x)+\sin ^{-1} x$ $\frac{d y}{d x}=-1+\frac{1}{\sqrt{1-x^{2}}}$ $\mathrm{x}=\frac{\sqrt{3}}{2}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=-1+2=1$