COMEDK · Maths · 23. Inverse Trigonometric Functions
If \(\cos ^{-1} x=\alpha,(0 < x < 1)\) and \(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}\), then \(\tan ^{-1}(2 x)=\)
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
Given that, \(\cos ^{-1} x=\alpha,(0 < x < 1)...(i)\) \(\Rightarrow \quad x=\cos \alpha\) Thus, \(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}\)…
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