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JEE Main 8-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 8-April-2025 Paper - Shift 24 Marks
MCQ
The value of $\cot ^{-1}\left(\frac{\sqrt{1+\tan ^{2}(2)}-1}{\tan (2)}\right)-\cot ^{-1}$ $\left(\frac{\sqrt{1+\tan ^{2}\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ is equal to
  • ✓
    $\pi-\frac{5}{4}$
  • B
    $\pi-\frac{3}{2}$
  • C
    $\pi+\frac{3}{2}$
  • D
    $\pi+\frac{5}{2}$
✓

Answer

Correct option: A.
$\pi-\frac{5}{4}$
(A) $\pi-\frac{5}{4}$
$\cot ^{-1}\left(\frac{|\sec 2|-1}{\tan 2}\right)-\cot ^{-1}\left(\frac{\left|\sec \frac{1}{2}\right|+1}{\tan \frac{1}{2}}\right)$
\begin{equation*}
\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}
\end{equation*}

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