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JEE Main 2002
Exam
JEE Main
Year
2002
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2002 · Mathematics · Inverse Trigonometric Functions

^ -1 ( α )= ^ -1 ( α )=x , then x=

cot⁡−1(cos⁡α)=tan⁡−1(cos⁡α)=x\cot ^{-1}(\sqrt{\cos \alpha})=\tan ^{-1}(\sqrt{\cos \alpha})=x, then sin⁡x=\sin x=

Options

  1. Atan⁡2(α2)\tan ^2\left(\frac{\alpha}{2}\right)
  2. Bcot⁡2(α2)\cot ^2\left(\frac{\alpha}{2}\right)
  3. Ctan⁡α\tan \alpha
  4. Dcot⁡(α2)\cot \left(\frac{\alpha}{2}\right)

Correct answer

A. tan⁡2(α2)\tan ^2\left(\frac{\alpha}{2}\right)

Step-by-step solution

cot⁡−1(cos⁡α)−tan⁡−1(cos⁡α)=x\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x tan⁡−1(1cos⁡α)−tan⁡−1(cos⁡α)=x⇒tan⁡−11cos⁡α−cos⁡α1+1cos⁡α⋅cos⁡α=x\tan ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}\right)-\tan ^{-1}(\sqrt{\cos \alpha})=x \Rightarrow \tan ^{-1} \frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}=x ⇒tan⁡−11−cos⁡α2cos⁡α=x⇒tan⁡x=1−cos⁡α2cos⁡α\Rightarrow \tan ^{-1} \frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}}=x \Rightarrow \tan x=\frac{1-\cos \alpha}{2 \sqrt{\cos \alpha}} OR cot⁡x=2cos⁡α1−cos⁡α\cot x=\frac{2 \sqrt{\cos \alpha}}{1-\cos \alpha} or cosec⁡x=1+cos⁡α1−cos⁡α\operatorname{cosec} x=\frac{1+\cos \alpha}{1-\cos \alpha} sin⁡x=1−cos⁡α1+cos⁡α=1−(1−2sin⁡2α/2)1+2cos⁡2α/2−1\sin x=\frac{1-\cos \alpha}{1+\cos \alpha}=\frac{1-\left(1-2 \sin ^2 \alpha / 2\right)}{1+2 \cos ^2 \alpha / 2-1} or sin⁡x=tan⁡2α2\sin x=\tan ^2 \frac{\alpha}{2}
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