Question

Prove the following:
$\frac{sin2\theta }{1-cos2\theta }=cot\theta or\frac{1-co{s}^2\theta }{1+cos2\theta }=\frac{ta{n}^2\theta }{2}$

Answer 1

Solution:-

• $\frac{\sin 2\theta }{1-\cos 2\theta }=\cot \theta$

L.H.S.-

$\frac{\sin 2\theta }{1-\cos 2\theta }$

Using identity:-

$\cos 2\theta =1-2{\sin }^2\theta \Rightarrow 1-\cos 2\theta =2{\sin }^2\theta$

$\sin 2\theta =2\sin \theta \cos \theta$

$\therefore \frac{\sin 2\theta }{1-\cos 2\theta }=\frac{2\sin \theta \cos \theta }{2{\sin }^2\theta }$

$\Rightarrow =\frac{\cos \theta }{\sin \theta }$

$\Rightarrow =\cot \theta (\because \cot \theta =\frac{\cos \theta }{\sin \theta })$

$=$R.H.S.

Hence proved.

• $\frac{1-{\cos }^2\theta }{1+\cos 2\theta }=\frac{tan\theta }{2}$
  

L.H.S.-

$\frac{1-{\cos }^2\theta }{1+\cos 2\theta }$

Using identity:-

$\cos 2\theta =2{\cos }^2\theta -1\Rightarrow 1+\cos 2\theta =2{\cos }^2\theta$

${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow 1-{\cos }^2\theta ={\sin }^2\theta$

$\therefore \frac{1-{\cos }^2\theta }{1+\cos 2\theta }=\frac{{\sin }^2\theta }{2{\cos }^2\theta }$

$\Rightarrow =\frac{1}{2}\left(\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)$

$\Rightarrow =\frac{1}{2}{\tan }^2\theta (\because \tan \theta =\frac{\sin \theta }{\cos \theta })$