If `sin^(-1)(a-a^2/3+a^3/9-...)+cos^(-1)(1+b+b^2+...)=pi/2` then find `a` and `b`

Options

A. `b = (2a -3)/(3a)`
B. `b = (3a -2)/(2a)`
C. `a = (3)/(2 - 3b)`
D. `a = (2)/(3 - 2b)`

Answer

A, C

Explanation

The given relation is possible when
`a - (a^(2))/(3) + (a^(3))/(9) + ... = 1 + b + b^(2) + ..`
Also, `-le a - (a^(2))/(3) + (a^(3))/(9) + ... Le 1`
and `-1 le 1 + b + b^(2) + ... Le 1`
`rArr |b| lt 1 and |a| lt 3 and (a)/(1 + (a)/(3)) = (1)/(1 -b)`
or `(3a)/(a + 3) = (1)/(1 -b)`
or `3a - 3ab = a + 3`
or `2a - 3ab = 3`
or `b = (2a -3)/(3a) and a = (3)/(2 - 3b)`