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If
\(\sin ^{-1}\left(\frac{2 \alpha}{1+\alpha^2}\right)+\sin ^{-1}\left(\frac{2 \beta}{1+\beta^2}\right)=2 \tan ^{-1} x \text {, }\)
then \(\mathrm{x}=\)
Options:
\(\sin ^{-1}\left(\frac{2 \alpha}{1+\alpha^2}\right)+\sin ^{-1}\left(\frac{2 \beta}{1+\beta^2}\right)=2 \tan ^{-1} x \text {, }\)
then \(\mathrm{x}=\)
Solution:
1970 Upvotes
Verified Answer
The correct answer is:
\(\frac{\alpha+\beta}{1-\alpha \beta}\)
\(\begin{array}{r}
\sin ^{-1}\left(\frac{2 \alpha}{1+\alpha^2}\right)+\sin ^{-1}\left(\frac{2 \beta}{1+\beta^2}\right)=2 \tan ^{-1} x \\
\Rightarrow \quad 2 \tan ^{-1} \alpha+2 \tan ^{-1} \beta=2 \tan ^{-1} x \\
\quad\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] \\
\Rightarrow \quad \tan ^{-1} \alpha+\tan ^{-1} \beta=2 \tan ^{-1} x \\
{\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]}
\end{array}\)
\(\begin{aligned} & \tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1} x \\ & \tan ^{-1}\left(\frac{\alpha+\beta}{1-\alpha \beta}\right)=\tan ^{-1} x \\ & \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\ & \Rightarrow \quad x=\frac{\alpha+\beta}{1-\alpha \beta}\end{aligned}\)
\sin ^{-1}\left(\frac{2 \alpha}{1+\alpha^2}\right)+\sin ^{-1}\left(\frac{2 \beta}{1+\beta^2}\right)=2 \tan ^{-1} x \\
\Rightarrow \quad 2 \tan ^{-1} \alpha+2 \tan ^{-1} \beta=2 \tan ^{-1} x \\
\quad\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] \\
\Rightarrow \quad \tan ^{-1} \alpha+\tan ^{-1} \beta=2 \tan ^{-1} x \\
{\left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]}
\end{array}\)
\(\begin{aligned} & \tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1} x \\ & \tan ^{-1}\left(\frac{\alpha+\beta}{1-\alpha \beta}\right)=\tan ^{-1} x \\ & \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\ & \Rightarrow \quad x=\frac{\alpha+\beta}{1-\alpha \beta}\end{aligned}\)
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