TS EAMCET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}\), then \(x\) is equal to
- A \(\frac{1}{2}\)
- B \(\frac{\sqrt{3}}{2}\)
- C \(-\frac{1}{2}\)
- D \(-\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
We have \(\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}\) Let \(A=\sin ^{-1} x\) \(\begin{array}{ll} \Rightarrow & x=\sin A, \cos A=\sqrt{1-x^2} \\ \text {and } & B=\cos ^{-1} x \\ \Rightarrow & x=\cos B, \sin B=\sqrt{1-x^2} \end{array}\) Now,…
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