Question

If ${\sin }^{-1}(a-\frac{a^2}{3}+\frac{a^3}{9}+...)+{\cos }^{-1}(1+b+b^2+...)=\frac{\pi }{2}$, then

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

Answer 1

Answer: (A), (C)

The correct options are
A $b=\frac{2a-3}{3a}$
C $a=\frac{3}{2-3b}$

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

$\Rightarrow si{n}^{-1}\left(\frac{a}{1-\left(\frac{-a}{3}\right)}\right)+co{s}^{-1}\left(\frac{1}{1-b}\right)$

$\Rightarrow si{n}^{-1}\left(\frac{3a}{a+3}\right)+co{s}^{-1}\left(\frac{1}{1-b}\right)=\frac{\pi }{2}$
Therefore
$\frac{3a}{a+3}=\frac{1}{1-b}$

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

$\Rightarrow b=\frac{2a-3}{3a}$ ...(i)

$\Rightarrow 3ab=2a-3$

$\Rightarrow 3=a(2-3b)$

$\Rightarrow \frac{3}{2-3b}=a$ ...(ii)