Question
If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$ for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals
A.
$$ \frac{1}{2}$$
B.
1
C.
$$ - \frac{1}{2}$$
D.
$$- 1$$
Answer :
1
Solution :
$$\eqalign{
& {\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2} \cr
& \Rightarrow \,\,{\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) \cr
& \Rightarrow \,\,{\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = {\cos ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) \cr
& \Rightarrow \,\,{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - ..... = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - ..... \cr} $$
On both sides we have G.P. of infinite terms.
$$\eqalign{
& \therefore \,\,\frac{{{x^2}}}{{1 - \left( {\frac{{ - {x^2}}}{2}} \right)}} = \frac{x}{{1 - \left( { - \frac{x}{2}} \right)}} \cr
& \Rightarrow \,\,\frac{{2{x^2}}}{{2 + {x^2}}} = \frac{{2x}}{{2 + x}} \cr
& \Rightarrow \,\,2x + {x^3} = 2{x^2} + {x^3} \cr
& \Rightarrow \,\,x\left( {x - 1} \right) = 0 \cr
& \Rightarrow \,\,x = 0,1\,\,{\text{but }}0 < \left| x \right| < \sqrt 2 \cr
& \Rightarrow \,\,x = 1. \cr} $$