Question
If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x},$ prove that $\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}.$
✓
Answer
Let: $\text{a}=\tan\text{z}$ $\text{b}=\tan\text{y}$ Then, $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}+\sin^{-1}\frac{2\text{b}}{1+\text{b}^2}=2\tan^{-1}\text{x}$$\Rightarrow\sin^{-1}\frac{2\tan\text{z}}{1+\tan^2\text{z}}+\sin^{-1}\frac{2\tan\text{y}}{1+\tan^2\text{y}}=2\tan^{-1}\text{x}$
$\Rightarrow\sin^{-1}(\sin2\text{z})+\sin^{-1}(\sin2\text{y})=2\tan^{-1}\text{x}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$ $\Rightarrow2\text{z}+2\text{y}=2\tan^{-1}\text{x}$ $\Rightarrow\tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\text{x}$ $[\because\ \text{a}=\tan\text{z}\text{ and }\text{b}=\tan\text{y}]$ $\Rightarrow\tan^{-1}\frac{\text{a}+\text{b}}{1-\text{ab}}=\tan^{-1}\text{x}$ $\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$ $\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}$
$\Rightarrow\sin^{-1}(\sin2\text{z})+\sin^{-1}(\sin2\text{y})=2\tan^{-1}\text{x}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$ $\Rightarrow2\text{z}+2\text{y}=2\tan^{-1}\text{x}$ $\Rightarrow\tan^{-1}\text{a}+\tan^{-1}\text{b}=\tan^{-1}\text{x}$ $[\because\ \text{a}=\tan\text{z}\text{ and }\text{b}=\tan\text{y}]$ $\Rightarrow\tan^{-1}\frac{\text{a}+\text{b}}{1-\text{ab}}=\tan^{-1}\text{x}$ $\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$ $\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{1-\text{ab}}$
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