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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ^ -1 x- ^ -1 x=(π)/(6) , then x is equal to

If sin⁡−1x−cos⁡−1x=π6\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}, then xx is equal to

Options

  1. A12\frac{1}{2}
  2. B32\frac{\sqrt{3}}{2}
  3. C−12-\frac{1}{2}
  4. D−32-\frac{\sqrt{3}}{2}

Correct answer

B. 32\frac{\sqrt{3}}{2}

Step-by-step solution

We have sin⁡−1x−cos⁡−1x=π6\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6} Let A=sin⁡−1xA=\sin ^{-1} x⇒x=sin⁡A,cos⁡A=1−x2and B=cos⁡−1x⇒x=cos⁡B,sin⁡B=1−x2\begin{array}{ll} \Rightarrow & x=\sin A, \cos A=\sqrt{1-x^2} \\ \text {and } & B=\cos ^{-1} x \\ \Rightarrow & x=\cos B, \sin B=\sqrt{1-x^2} \end{array}Now,Now,sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B=x\cdotx−1−x2⋅1−x2=x2−1+x2sin⁡(π6)=2x2−1⇒2x2−1=12⇒2x2=32\Rightarrowx2=34⇒x=32\begin{aligned} \sin (A-B) & =\sin A \cos B-\cos A \sin B \\ & =x \cdotx-\sqrt{1-x^2} \cdot \sqrt{1-x^2}\\ & =x²-1+x² \\ \sin \left(\frac{\pi}{6}\right)& =2 x²-1 \\ \Rightarrow 2 x²-1 & =\frac{1}{2} \Rightarrow2 x²=\frac{3}{2} \Rightarrowx^2=\frac{3}{4}\\ \Rightarrow \quad x & =\frac{\sqrt{3}}{2} \end{aligned}$$
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