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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

The value of ^ -1 ( 1+ ^2(2) -1 (2) )- ^ -1 ( 1+ ^2((1)/(2)) +1...

The value of cot⁡−1(1+tan⁡2(2)−1tan⁡(2))−cot⁡−1\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 }

Options

  1. Aπ−54\pi-\frac{5}{4}
  2. Bπ−32\pi-\frac{3}{2}
  3. Cπ+32\pi+\frac{3}{2}
  4. Dπ+52\pi+\frac{5}{2}

Correct answer

A. π−54\pi-\frac{5}{4}

Step-by-step solution

cot⁡−1(∣sec⁡2∣−1tan⁡2)−cot⁡−1(∣sec⁡12∣+1tan⁡12)\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)
=cot⁡−1(−1−cos⁡2sin⁡2)−cot⁡−1(1+cos⁡12sin⁡12)=π−cot⁡−1(cot⁡1)−cot⁡−1(cot⁡14)=π−1−14=π−54\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}
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