TS EAMCET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\), then \(\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=\)
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ = & \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \Rightarrow \sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} 2 x \end{aligned}\) Take sine on both sides…
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