Q36 / 75 · JEE Main 2025 (8 Apr Shift 2) · Mathematics
The value of $\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2(2)}-1}{\tan (2)}\right)-\cot ^{-1}$
$\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right) \text { is equal to }$
$\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right) \text { is equal to }$
- Correct answerA$\pi-\frac{5}{4}$
- B$\pi-\frac{3}{2}$
- C$\pi+\frac{3}{2}$
- D$\pi+\frac{5}{2}$
Solution
$\cot ^{-1}\left(\frac{|\sec 2|-1}{\tan 2}\right)-\cot ^{-1}\left(\frac{\left|\sec \frac{1}{2}\right|+1}{\tan \frac{1}{2}}\right)$
$\begin{aligned} & =\cot ^{-1}\left(\frac{-1-\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right) \\ & =\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right) \\ & =\pi-1-\frac{1}{4}=\pi-\frac{5}{4}\end{aligned}$
$\begin{aligned} & =\cot ^{-1}\left(\frac{-1-\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right) \\ & =\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right) \\ & =\pi-1-\frac{1}{4}=\pi-\frac{5}{4}\end{aligned}$
Hence, the correct answer is option: 'A'.