COMEDK · Maths · 23. Inverse Trigonometric Functions
If \(\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right)\) \(=\tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then the value of \(x\) is equal to
- A \(\frac{p+q}{1+p q}\)
- B \(\frac{p-q}{1-p q}\)
- C \(\frac{p-q}{p q-1}\)
- D \(\frac{p-q}{1+p q}\)
Answer & Solution
Correct Answer
(D) \(\frac{p-q}{1+p q}\)
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