Question

Let $f\left(x\right)=\cos \left({\tan }^{-1}2x\right)-\sin \left\{{\tan }^{-1}\left(\frac{1}{2x+1}\right)\right\}$ and $a=\cos \left({\tan }^{-1}\right(\sin \left({\cot }^{-1}2x\right)\left)\right)$ and $b=\cos (\frac{\pi }{2}+co{s}^{-1}2x)$. If $f\left(x\right)=0$ , then $b=$

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $-\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$-\frac{\sqrt{3}}{2}$

Here, $f\left(x\right)=0$
$f\left(x\right)=cos\left(ta{n}^{-1}2x\right)-sin\left\{ta{n}^{-1}\left(\frac{1}{2x+1}\right)\right\}=0$
$\cos ({\tan }^{-1}2x)=\sin \left[{\tan }^{-1}\left(\frac{1}{2x+1}\right)\right]$
Lets assume

$A=\sin \left[{\tan }^{-1}\left(\frac{1}{2x+1}\right)\right]$
$=\sin \left[{\sin }^{-1}\left(\frac{1}{\sqrt{4x^2+4x+2}}\right)\right]$
$=\left(\frac{1}{\sqrt{4x^2+4x+2}}\right)$ (1)

$B=\cos \left({\tan }^{-1}2x\right)$
$=\cos \left[{\cos }^{-1}\left(\frac{1}{\sqrt{4x^2+1}}\right)\right]$
$=\left(\frac{1}{\sqrt{4x^2+1}}\right)$ (2)

As $f\left(x\right)=0$. Equate A and B
$\left(\frac{1}{\sqrt{4x^2+4x+2}}\right)=\left(\frac{1}{\sqrt{4x^2+1}}\right)$

Squaring both sides
$1+4x^2=4x^2+4x+2$
$4x+1=0$
$x=-1/4$
$b=\cos (\frac{\pi }{2}+{\cos }^{-1}2x)$

$=-\sin \left({\sin }^{-1}\right(\sqrt{1-4x^2}\left)\right)$
$=-\sqrt{1-4x^2}$
$=\frac{-\sqrt{3}}{2}$