Question

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

• A. $\frac{1}{2}\tan \theta$
• B. $\frac{1}{2}\cot \theta$
• C. $\tan \theta$
• D. $\cot \theta$

Answer 1

The correct option is D $\cot \theta$

The value of $\frac{1+\sin 2\theta +\cos 2\theta }{1+\sin 2\theta -\cos 2\theta }$

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

$=\frac{2\cos \theta [\cos \theta +\sin \theta ]}{2\sin \theta [\sin \theta +\cos \theta ]}=\frac{\cos \theta }{\sin \theta }=\cot \theta$