Question
$${\sin ^{ - 1}}\left( {a - \frac{{{a^2}}}{3} + \frac{{{a^3}}}{9} + .....} \right) + {\cos ^{ - 1}}\left( {1 + b + {b^2} + .....} \right) = \frac{\pi }{2}{\text{ when}}$$
A.
$$a = - 3\,{\text{and}}\,b = 1$$
B.
$$a = 1\,{\text{and}}\,b = - \frac{1}{3}$$
C.
$$a = \frac{1}{6}\,{\text{and}}\,b = \frac{1}{2}$$
D.
None of these
Answer :
$$a = 1\,{\text{and}}\,b = - \frac{1}{3}$$
Solution :
The given relation is possible when
$$\eqalign{
& a - \frac{{{a^2}}}{3} + \frac{{{a^3}}}{9} + ..... = 1 + b + {b^2} + ..... \cr
& {\text{Also }} - 1 \leqslant a - \frac{{{a^2}}}{3} + \frac{{{a^3}}}{9} + ..... \leqslant 1 \cr
& {\text{and}}\, - 1 \leqslant 1 + b + {b^2} + ..... \leqslant 1 \cr
& \Rightarrow \left| b \right| < 1 \cr
& \Rightarrow \left| {a} \right| < 3 \,{\text{ and }}\,\frac{a}{{1 + \frac{a}{3}}} = \frac{1}{{1 - b}} \cr
& \Rightarrow \frac{{3a}}{{a + 3}} = \frac{1}{{1 - b}} \cr} $$
There are infinitely many solutions. But in the given options, it is satisified only when
$$a = 1\,\,{\text{and }}\,b = - \frac{1}{3}.$$