Question

(a) Prove that $\sin [{tan}^{-1}\frac{1-x^2}{2x}+{\cos }^{-1}\frac{1-x^2}{1+x^2}]=1$.

(b) If ${\sin }^{-1}(x-\frac{x^2}{2}+\frac{x^3}{4}-.....)+{\cos }^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-.......)=\frac{\pi }{2}$
for $0<|x|<\sqrt{2}$, then x equals

• A. $\frac{1}{2}$
• B. $1$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (B)

$1$

(a) $\sin [{\tan }^{-1}\frac{1-x^2}{2x}+{\cos }^{-1}\frac{1-x^2}{1+x^2}]=1$

Put $x=\tan \theta$

L.H.S.$=\sin [{\tan }^{-1}\frac{1-{\tan }^2\theta }{2\tan \theta }+{\cos }^{-1}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }]$

$=\sin [{\tan }^{-1}cot2\theta +{\cos }^{-1}\cos 2\theta ]$

$=\sin \left[{\tan }^{-1}\tan \right(\pi /2-2\theta )+2\theta ]$

$=\sin [\pi /2-2\theta +2\theta ]=\sin \left(\pi /2\right)=1$.

Hence proved

(b)

Given ${\sin }^{-1}A+{\cos }^{-1}B=\frac{\pi }{2}$.........(1)

But ${\sin }^{-1}A+{\cos }^{-1}A=\frac{\pi }{2}$........(2)

From (1) and (2),we conclude that $A=B$

$\therefore$ $x-\frac{x^2}{2}+\frac{x^3}{4}....=x^2-\frac{x^4}{2}+\frac{x^6}{4}.....$

or $\frac{x}{1-(-\frac{x}{2})}=\frac{x^2}{1--\left(\frac{x^2}{2}\right)}$ by $S_{\infty }$ of a G.P.

or $x(2+x^2)=x^2(2+x)$

or $2x(x-1)=0$

$\therefore$ $x=1$ as $x\ne 0$