Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 23 January, Morning Shift — Question 14

If π2≤x≤3π4\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}, then cos⁡−1(1213cos⁡x+513sin⁡x)\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)

is equal to

  1. Option A:

    x−tan⁡−143x-\tan ^{-1} \frac{4}{3}

  2. Option B:

    x−tan⁡−1512x-\tan ^{-1} \frac{5}{12}

    Correct
  3. Option C:

    x+tan⁡−145x+\tan ^{-1} \frac{4}{5}

  4. Option D:

    x+tan⁡−1512x+\tan ^{-1} \frac{5}{12}

Answer: B

Step-by-step solution

π2≤x≤3π4\frac{\pi}{2} \leq \mathrm{x} \leq \frac{3 \pi}{4}

cos⁡−1(1213cos⁡x+513sin⁡x)\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)

cos⁡−1(cos⁡xcos⁡α+sin⁡xsin⁡α)\cos ^{-1}(\cos x \cos \alpha+\sin x \sin \alpha)

cos⁡−1(cos⁡(x−α))\cos ^{-1}(\cos (x-\alpha))

⇒x−α\Rightarrow \mathrm{x}-\alpha because x−α∈(−π2,π2)\mathrm{x}-\alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

⇒x−tan⁡−1512\Rightarrow \mathrm{x}-\tan ^{-1} \frac{5}{12}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions
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