MHT CET · Maths · Inverse Trigonometric Functions
If \(A=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right)\) and \(B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\), where \(x \in(0,1)\), then \(A-B=\)
- A \(\frac{\pi}{4}\)
- B \(4 \tan ^{-1} x\)
- C \(\tan ^{-1} x\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Let \(x=\tan \theta\) then
\(\begin{aligned} & A=2 \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \text { and } B=\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow A=2\left\{\frac{\pi}{4}+\theta\right\} \text { and } B=2 \theta \\ & \Rightarrow A-B=\frac{\pi}{2}\end{aligned}\)
\(\begin{aligned} & A=2 \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \text { and } B=\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow A=2\left\{\frac{\pi}{4}+\theta\right\} \text { and } B=2 \theta \\ & \Rightarrow A-B=\frac{\pi}{2}\end{aligned}\)
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