JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 16

First term. $\dfrac{\frac34-\tan x}{1+\frac34\tan x}=\tan(\theta_0-x)$ with $\tan\theta_0=\frac34$, so the term is $\theta_0-x$ near $x=\frac{\sqrt3}2$. Derivative $-1$.

Second term. With $x=\sin\phi$: $\dfrac{\sin\phi}{1+\cos\phi}=\tan\frac\phi2$, so the term is $\phi=\sin^{-1}x$. Derivative $\dfrac1{\sqrt{1-x^2}}=2$ at $x=\frac{\sqrt3}2$.

Total $-1+2=1$.

(C)