MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}\) is
- A \(\frac{x+y}{1-x y}\)
- B \(\frac{x-y}{1+x y}\)
- C \(\frac{x-y}{1-x y}\)
- D \(\frac{x+y}{1+x y}\)
Answer & Solution
Correct Answer
(A) \(\frac{x+y}{1-x y}\)
Step-by-step Solution
Detailed explanation
Let \(x=\tan \theta\) and \(y=\tan \phi\)
\(\begin{aligned}
& \Rightarrow \tan \left\{\frac{1}{2} \sin ^{-1} \sin 2 \theta+\frac{1}{2} \cos ^{-1} \cos 2 \phi\right\} \\
& =\tan (\theta+\phi) \\
& =\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}=\frac{x+y}{1-x y}
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow \tan \left\{\frac{1}{2} \sin ^{-1} \sin 2 \theta+\frac{1}{2} \cos ^{-1} \cos 2 \phi\right\} \\
& =\tan (\theta+\phi) \\
& =\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}=\frac{x+y}{1-x y}
\end{aligned}\)
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