MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]\) is
- A \(0\)
- B \(1\)
- C \(-1\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
Let \(x = \tan\theta\). \(\tan^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{1-\tan^2\theta}{2\tan\theta}\right) = \tan^{-1}(\cot(2\theta)) = \tan^{-1}\left(\tan\left(\frac{\pi}{2}-2\theta\right)\right) = \frac{\pi}{2}-2\theta\)
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