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Mathematics · Inverse Trigonometric Functions

JEE Advanced 2024 — Paper 2 — Question 1

Considering only the principal values of the inverse trigonometric functions, the value of tan⁡(sin⁡−1(35)−2cos⁡−1(25))\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) is

  1. Option A:

    724\frac{7}{24}

  2. Option B:

    −724\frac{-7}{24}

    Correct
  3. Option C:

    −524\frac{-5}{24}

  4. Option D:

    524\frac{5}{24}

Answer: B

Step-by-step solution

tan⁡(sin⁡−135−2cos⁡−125)=tan⁡(tan⁡−134−2tan⁡−112)\quad \tan \left(\sin ^{-1} \frac{3}{5}-2 \cos ^{-1} \frac{2}{\sqrt{5}}\right)=\tan \left(\tan ^{-1} \frac{3}{4}-2 \tan ^{-1} \frac{1}{2}\right)

=tan⁡(tan⁡−134−tan⁡−12×121−(12)2)=tan⁡(tan⁡−134−tan⁡−143)=\tan \left(\tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right)=\tan \left(\tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{4}{3}\right)

=tan⁡(tan⁡−1(−724))=−724=\tan \left(\tan ^{-1}\left(\frac{-7}{24}\right)\right)=\frac{-7}{24}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions