Question

If $si{n}^{-1}(a-\frac{a^2}{3}+\frac{a^3}{9}-......\infty )+co{s}^{-1}(1+b+b^2+....\infty )=\frac{\pi }{2}$ then

• A. $a=-3,b=1$
• B. $a=1,b=\frac{-1}{3}$
• C. $a=\frac{1}{6},b=\frac{1}{2}$
• D. $a=\frac{1}{6},b=\frac{1}{3}$

Answer 1

The correct option is B $a=1,b=\frac{-1}{3}$

$\frac{a}{1+\frac{a}{3}}=\frac{1}{1-b}\Rightarrow \frac{3a}{a+3}=\frac{1}{1-b}$
There are infinitely many solutions but option ‘b’ satisfies.