Question

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

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

Answer 1

Answer: (A)

$a=-3$ and $b=1$

$\begin{array}{c}{\sin }^{-1}\left(\frac{a}{1-\left(-\frac{a}{3}\right)}\right)+{\cos }^{-1}\left(\frac{1}{1-b}\right)=\frac{\pi }{2}\\ \left[\because AngleareinG.PthensumofiniteG.P\right]\\ {\sin }^{-1}\left(\frac{a}{1+\frac{a}{3}}\right)+{\cos }^{-1}\left(\frac{1}{1-b}\right)=\frac{\pi }{2}is\left(\frac{a}{1-r}\right)\\ i{n}^{-1}\left(\frac{3a}{a+3}\right)+{\cos }^{-1}\left(\frac{1}{1-b}\right)=\frac{\pi }{4}+\frac{\pi }{4}.........\left(i\right)\\ byinversepropertyifangleof\sin and\cos ineissamethen\\ {Sin}^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\\ Bycompearing\\ {Sin}^{-1}\left(\frac{3a}{a+3}\right)=\frac{\pi }{4}\\ Sin\frac{\pi }{4}=\frac{3a}{a+3}\\ \frac{1}{\sqrt{2}}=\frac{3a}{a+3}\\ a+3=3\sqrt{2a}\\ 3=a(3\sqrt{2-1}\\ a=\frac{3}{3\sqrt{2-1}}\\ {\cos }^{-1}\left(\frac{1}{1-b}\right)=\frac{\pi }{4}\\ \cos \frac{\pi }{4}=\frac{1}{1-b}\\ \frac{1}{\sqrt{2}}=\frac{1}{1-b}\\ \sqrt{2}=1-b\\ b=1-\sqrt{2}\end{array}$