Question

${\sin }^{-1}(a-\frac{a^2}{3}+\frac{a^3}{9}\cdots \infty )+{\cos }^{-1}(1+b+b^2+b^3+\cdots \infty )=\frac{\pi }{2}$ , when

• A. $a=1,b=-\frac{1}{3}$
• B. $a=-\frac{1}{6},b=\frac{1}{2}$
• C. $a=\frac{1}{6},b=\frac{1}{2}$
• D. None of these

Answer 1

Answer: (D)

None of these

$a-\frac{a^2}{3}+\frac{a^3}{9}...\infty$ is an infinite G.P with a common ration of $\frac{-a}{3}$.

Hence
$S_a=\frac{a}{1+\frac{a}{3}}$ $=\frac{3a}{a+3}$ ...(i)

Similarly $S_b=1+b+b^2...\infty$.
$S_b=\frac{1}{1-b}$ ...(ii)

Now $si{n}^{-1}\left(A\right)+co{s}^{-1}\left(B\right)=\frac{\pi }{2}$ implies $A=B$, then

$i=ii$

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

$3a-3ab=a+3$

$2a-3ab=3$

$2a-3=3ab$

$b=\frac{2a-3}{3a}$

Substituting $b=\frac{1}{2}$, we get

$\frac{3a}{2}-2a=-3$

$\frac{a}{2}=3$

$a=6$.

However $|a|<1$ and $|b|<1$ for the above infinite G.P.

Hence answer is none of the above.