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# Solve for x $\sin^{-1}\frac{2\alpha}{1+\alpha^2}+ \sin^{-1}\frac{2\beta}{1+\beta^2}=2\tan^{-1}x.$

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Toolbox:
• $sin2A= \large\frac{2tanA}{1=tan^2A}$
• $tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}$
put $\alpha = tanA \: and \: \beta = tanB$
$\Rightarrow\: A=tan^{-1}\alpha \; \: B=tan^{-1}\beta$
$\large\frac{2\alpha}{1+\alpha^2}=\large\frac{2tanA}{1+tan^2A}=sin2A$
$\large\frac{2\beta}{1+\beta^2}=\large\frac{2tanB}{1+tan^2B}=sin2B$

Substituting in L.H.S. we get
$sin^{-1} \large\frac{2\alpha}{1+\alpha^2}+sin^{-1}\large\frac{2\beta}{1+\beta^2}$
$sin^{-1} sin2A+sin^{-1}\: sin2B$
$=2A+2B=2(A+B)=2 \bigg[ tan^{-1}\alpha+tan^{-1}\beta \bigg]$

Substituting L.H.S. in the given equation , we get
$\Rightarrow 2(tan^{-1}\alpha + tan^{-1}\beta) =2 tan^{-1}x$
$\Rightarrow tan^{-1}\alpha + tan^{-1}\beta = tan^{-1}x$
By taking $\alpha$ in the place of x and $\beta$ in the place of y in the above forfula of $tan^{-1}x+tan^{-1}y$we get
$\Rightarrow\: tan^{-1}\alpha + tan^{-1}\beta= tan^{-1}\large\frac{\alpha+\beta}{1-\alpha\beta}=tan^{-1}x$
or $\Rightarrow\: x=\large\frac{\alpha+\beta}{1-\alpha\beta}$

answered Feb 19, 2013
edited Mar 19, 2013