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# Prove that $\cos\bigg\{2\cot^{-1}\sqrt{\large\frac{1-x}{1+x}}\bigg\}+x=0$

Can you answer this question?

Toolbox:
• put $x=cos2\theta \Rightarrow 2\theta=cos^{-1}x$
• $1-cos2\theta=2\: sin^2\theta$
• $1+cos2\theta = 2cos^2\theta$
• $cot^{-1}\: tan\theta = \large\frac{\pi}{2}-\theta$
• $cos(\pi-x)=-cosx$
By taking $x=cos2\theta,\:\:\sqrt{\large\frac{1-x}{1+x}}=\sqrt{\large\frac{1-cos2\theta}{1+cos2\theta}}$
$=\sqrt{\large\frac{2sin^2\theta}{2cos^2\theta}}=\frac{sin\theta}{cos\theta}=tan\theta$
$\Rightarrow\:cot^{-1}\sqrt{\large\frac{1-x}{1+x}}=cot^{-1}tan\theta$

Substituting the value of x in the given eqn. we get
$cos \{ 2 \: cot^{-1}\: tan\theta \} + cos2\theta$

$\Rightarrow\: cos \{ 2 ( \large\frac{\pi}{2} - \theta ) \} + cos2\theta$
$=cos(\pi-2\theta)+cos2\theta$
$= -cos2\theta+cos2\theta=0$

answered Feb 20, 2013
edited Mar 19, 2013