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Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 29 January, Shift 1 — Question 10

If α,−π2<α<π2\alpha ,-\frac{\pi }{2}<\alpha <\frac{\pi }{2} is the solution of 4cos⁡θ+5sin⁡θ=1,4\cos \theta +5\sin \theta =1, then the value of tan⁡α\tan \alpha is

  1. Option A:
    10−106\frac{10-\sqrt{10}}{6}
  2. Option B:
    10−1012\frac{10-\sqrt{10}}{12}
  3. Option C:
    10−1012\frac{\sqrt{10}-10}{12}
    Correct
  4. Option D:
    10−106\frac{\sqrt{10}-10}{6}

Answer: C

Step-by-step solution

4+5tan⁡θ=sec⁡θ4+5 \tan \theta=\sec \theta

Squaring : 24tan⁡2θ+40tan⁡θ+15=024 \tan ^{2} \theta+40 \tan \theta+15=0

tan⁡θ=−10±1012\tan \theta=\frac{-10 \pm \sqrt{10}}{12} and tan⁡θ=−(10+1012)\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right) is Rejected.

(3) is correct.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Introduction to Trigonometry