MCQ
यदि $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^2}\right)=2 \tan ^{-1} x$, तब $x=$
- A$\frac{a-b}{1+a b}$
- B$\frac{b}{1+a b}$
- C$\frac{b}{1-a b}$
- ✓$\frac{a+b}{1-a b}$
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