Question

The number of solutions of ${\sin }^{-1}(1+b+b^2+\cdots \infty )+{\cos }^{-1}(a-\frac{a^2}{3}+\frac{a^2}{9}\cdots \infty )=\frac{\pi }{2}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $\infty$

Answer 1

Answer: (D)

$\infty$

Given : ${\sin }^{-1}(1+b+b^2+\cdots \infty )+{\cos }^{-1}(a-\frac{a^2}{3}+\frac{a^2}{9}\cdots \infty )=\frac{\pi }{2}$
The given equation will be valid if $-1\le 1+b+b^2+\cdots \infty \le 1$
and $a-\frac{a^2}{3}+\frac{a^3}{9}-\cdots \infty ϵ[-1,1]$
${\sin }^{-1}(1+b+b^2+\cdots \infty )={\sin }^{-1}(a-\frac{a^2}{3}+\frac{a^2}{9}\cdots \infty )$
Also $1+b+b^2+\cdots \infty =a-\frac{a^2}{3}+\frac{a^3}{9}\cdots \infty$
(Infinite GP. series)
$\Rightarrow \frac{1}{1-b}=\frac{a}{1+\frac{a}{3}}$
$\Rightarrow \frac{1}{1-b}=\frac{3a}{a+3}$
which is true for infinite number of values of $a$ and $b$
So, $\exists$ infinite solutions of the equation.