Question

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

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

Answer 1

The correct option is D $\infty$

As ${\sin }^{-1}\theta$ are defined for $\theta \in [-1,1]$

$\therefore$ For ${\sin }^{-1}(1+b+b^3+...\infty )$

$\left|b\right|<1$

Hence $1,b,b^2,...\infty$ is $G.P.$ of common ratio $b$

$\therefore 1+b+b^2+...+\infty =\frac{1}{1-b}$

Similarly for ${\cos }^{-1}(a-\frac{a^2}{3}+\frac{a^3}{9}+...\infty )$

$\left|\frac{a}{3}\right|<1$

Hence $1,-\frac{a^2}{3},\frac{a^3}{9},...\infty$ is $G.P.$ of common ration $\left(\frac{-a}{3}\right)$

$\therefore a-\frac{a^2}{3}+\frac{a^3}{9}+...\infty =\frac{a}{1+\frac{a}{3}}=\frac{3a}{3+a}$

Now, as ${\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\frac{\pi }{2}$

$\therefore \frac{1}{1-b}=\frac{3a}{3+a}$

$\Rightarrow 3+a=3a-3ab$

$\Rightarrow 2a-3=3ab$

This has infinite solution.