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Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 8 April, Evening Shift — Question 24

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
  1. Option A:

    π−32\pi-\frac{3}{2}

  2. Option B:

    π−54\pi-\frac{5}{4}

    Correct
  3. Option C:

    π+32\pi+\frac{3}{2}

  4. Option D:

    π+52\pi+\frac{5}{2}

Answer: B

Step-by-step solution

cot⁡−1(1+tan⁡22−1tan⁡2)−cot⁡−1(1+tan⁡2(12)+1tan⁡12)\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 \frac{1}{2}}\right)

=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 \left(\frac{1}{2}\right)\right|+1}{\tan \frac{1}{2}}\right)

=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{\sec \frac{1}{2}+1}{\tan \frac{1}{2}}\right)

=π−cot⁡−1(1+cos⁡2sin⁡2)−cot⁡−1(1+cos⁡12sin⁡12)=\pi-\cot ^{-1}\left(\frac{1+\cos 2}{\sin 2}\right)-\cot ^{-1}\left(\frac{1+\cos \frac{1}{2}}{\sin \frac{1}{2}}\right)

=π−cot⁡−1(2cos⁡212sin⁡1⋅cos⁡1)−cot⁡−1(2cos⁡2142sin⁡14⋅cos⁡14)=\pi-\cot ^{-1}\left(\frac{2 \cos ^{2} 1}{2 \sin 1 \cdot \cos 1}\right)-\cot ^{-1}\left(\frac{2 \cos ^{2} \frac{1}{4}}{2 \sin \frac{1}{4} \cdot \cos \frac{1}{4}}\right)

=π−cot⁡−1(cot⁡1)−cot⁡−1(cot⁡14)=\pi-\cot ^{-1}(\cot 1)-\cot ^{-1}\left(\cot \frac{1}{4}\right)

=π−1−14=\pi-1-\frac{1}{4}

=π−54=\pi-\frac{5}{4}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Equations, Inequations and Identitites involving ITFs