Question

Which of the following identities, wherever defined, hold(s) good?

• A. $\cot \alpha -\tan \alpha =2\cot 2\alpha$
• B. $\tan ({45}^{\circ }+\alpha )-\tan ({45}^{\circ }-\alpha )=2cosec2\alpha$
• C. $\tan ({45}^{\circ }+\alpha )+\tan ({45}^{\circ }-\alpha )=2\sec 2\alpha$
• D. $\tan \alpha +\cot \alpha =2\tan 2\alpha$

Answer 1

Answer: (A), (C)

$\cot \alpha -\tan \alpha =2\cot 2\alpha$
$\tan ({45}^{\circ }+\alpha )+\tan ({45}^{\circ }-\alpha )=2\sec 2\alpha$

Option A is true as
$\cot \alpha -\tan \alpha =\frac{\cos \alpha }{\sin \alpha }-\frac{\sin \alpha }{\cos \alpha }=\frac{{\cos }^2\alpha -{\sin }^2\alpha }{\sin \alpha \cos \alpha }=\frac{\cos 2\alpha }{\sin \alpha \cos \alpha }=\frac{2\cos 2\alpha }{\sin 2\alpha }=2\cot 2\alpha$
Option B is wrong as
$\tan (45+\alpha )-\tan (45-\alpha )=\frac{\tan 45+\tan \alpha }{1-\tan 45\tan \alpha }-\frac{\tan 45-\tan \alpha }{1+\tan 45\tan \alpha }=\frac{1+\tan \alpha }{1-\tan \alpha }-\frac{1-\tan \alpha }{1+\tan \alpha }=\frac{{(1+\tan \alpha )}^2-{(1-\tan \alpha )}^2}{1-{\tan }^2\alpha }=\frac{1+{\tan }^2\alpha +2\tan \alpha -1-{\tan }^2\alpha +2\tan \alpha }{1-({\sec }^2\alpha -1)}=\frac{4\tan \alpha }{2-{\sec }^2\alpha }=\frac{4\sin \alpha \cos \alpha }{2{\cos }^2\alpha -1}=2\tan 2\alpha$
Option C is true as
$\tan (45+\alpha )+\tan (45-\alpha )=\frac{\tan 45+\tan \alpha }{1-\tan 45\tan \alpha }+\frac{\tan 45-\tan \alpha }{1+\tan 45\tan \alpha }=\frac{1+\tan \alpha }{1-\tan \alpha }+\frac{1-\tan \alpha }{1+\tan \alpha }=\frac{{(1+\tan \alpha )}^2+{(1-\tan \alpha )}^2}{1-{\tan }^2\alpha }=\frac{1+{\tan }^2\alpha +2\tan \alpha +1+{\tan }^2\alpha -2\tan \alpha }{1-({\sec }^2\alpha -1)}=\frac{2{\sec }^2\alpha }{2-{\sec }^2\alpha }=\frac{2}{2{\cos }^2\alpha -1}=\frac{2}{{\cos }^2\alpha -{\sin }^2\alpha }=\frac{2}{\cos 2\alpha }=2\sec \alpha$
Option D is wrong as
$\tan \alpha +\cot \alpha =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }=\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }=\frac{2}{\sin 2\alpha }=2cosec2\alpha$