If $\sin^{-1}\frac{2a}{1+a^2} +\cos^{-1}\frac{1-a^2}{1+a^2}=\tan^{-1}\frac{2x}{1-x^2},where\; a,x,\in\;[0,1],$then the value of x is

$(a)\;0\qquad(b)\;\large\frac{a}{2}\qquad(c)\;a\qquad(d)\;\large\frac{2a}{1-a^{2}}$

Toolbox:
• Put $$a=tanA \Rightarrow A = tan^{-1}a$$
• $$\large\frac{2tanA}{1+tan^2A}=sin2A$$
• $$\large\frac{1-tan^2A}{1=tan^2A}=cos2A$$
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2},\:\:\:|x|<1$$
Ans (D) $$\large\frac{2a}{1-a^2}$$

By taking a=tanA, $$tan^{-1}a=A$$
$$\large\frac{2a}{1+a^2}=\large\frac{2tanA}{1+tan^2A}=sin2A\:\:and$$
$$\large\frac{1-a^2}{1+a^2}= \large\frac{1-tan^2A}{1=tan^2A}=cos2A$$
$$sin^{-1}\large\frac{2a}{1+a^2}=sin^{-1}sin2A=2A$$
$$cos^{-1}\large\frac{1-a^2}{1+a^2}=cos^{-1}cos2A=2A$$
L.H.S. becomes
$$sin^{-1}\large\frac{2a}{1+a^2}+cos^{-1}\large\frac{1-a^2}{1+a^2}$$
$$=2A+2A=4A=4tan^{-1}a$$
Taking R.H.S., $$\tan^{-1}\large\frac{2x}{1-x^2}=2tan^{-1}x$$
Substituting the values of L.H.S. and R.H.S. in the given equation, we get
$$4tan^{-1}a=tan^{-1}\large \frac{2x}{1-x^2} =2tan^{-1}x$$

$$\Rightarrow 2tan^{-1}a=tan^{-1}x$$
By using the above formula of $$2tan^{-1}x$$
$$\Rightarrow\:2tan^{-1}a= tan^{-1} \large\frac{2a}{1-a^2}$$
$$\Rightarrow\:tan^{-1} \large\frac{2a}{1-a^2}=tan^{-1}x$$
$$\Rightarrow\: \large\frac{2a}{1-a^2}=x$$

edited Mar 16, 2013