JEE Main 2025 Apr 8 Shift 2, Mathematics Q18: Inverse Trigonometric Functions

Use the identity $\sqrt{1+{\tan }^{2}x}=|\sec x|$; for the angles here, take $|\sec 2|$ (with $2$ radians in the second quadrant, so $\sec 2 < 0$, giving $|\sec 2| = -\sec 2$) and $\sec(1/2) > 0$.

For the first argument: $\frac{|\sec 2|-1}{\tan 2}=\frac{-\sec 2-1}{\tan 2}=\frac{-1-\cos 2}{\sin 2}$ after multiplying numerator and denominator by $\cos 2$.

Using $1+\cos 2=2{\cos }^{2}1$ and $\sin 2=2\sin 1\cos 1$: this equals $-\cot 1$.

So ${\cot }^{-1}(-\cot 1)=\pi -{\cot }^{-1}(\cot 1)=\pi -1$.

For the second argument: $\frac{\sec (1/2)+1}{\tan (1/2)}=\frac{1+\cos (1/2)}{\sin (1/2)}$ after the same simplification. Using $1+\cos (1/2)=2{\cos }^2(1/4)$ and $\sin (1/2)=2\sin (1/4)\cos (1/4)$, this equals $\cot (1/4)$.

So ${\cot }^{-1}(\cot (1/4))=1/4$.

Difference: $(\pi -1)-\frac{1}{4}=\pi -\frac{5}{4}$, matching option (1).