If x=1−cosθ1+cosθ then 2x1−x2=

x=1−cosθ1+cosθ =2sin2θ22cos2θ2 ⇒x=sinθ2cosθ2=tanθ2 ⇒2x1−x2=2tanθ21−tan2θ2 =tanθ

If x=1−cosθ1+cosθ then 2x1−x2=

x=1−cosθ1+cosθ =2sin2θ22cos2θ2 ⇒x=sinθ2cosθ2=tanθ2 ⇒2x1−x2=2tanθ21−tan2θ2 =tanθ

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If $x = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}}$ then $\frac{2 x}{1 - x^{2}} =$

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\cos θ

\tan θ

\sin θ

\cot θ

answer is C.

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Detailed Solution

x = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}}

$= \sqrt{\frac{2 \sin^{2} \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}}$

$\Rightarrow x = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}} = \tan \frac{θ}{2}$

$\Rightarrow \frac{2 x}{1 - x^{2}} = \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}}$

$= \tan θ$

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If $x = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}}$ then $\frac{2 x}{1 - x^{2}} =$