Question

prove
$\cot 2 a + \tan a = \cosec 2a$

Answer 1

We will use one identity:

$\boxed{\cos A \cos B + \sin A \sin B = \cos (A-B)}$


To Prove:

$\cot 2a + \tan a = cosec \, 2a$

Consider:

$\mathbb{LHS} \\ \\ \\ = \cot 2a + \tan a \\ \\ \\ = \frac{\cos 2a}{\sin 2a} +\frac{\sin a}{\cos a} \\ \\ \\ = \frac{\cos 2a \cos a \, + \, \sin 2a \sin a}{\sin 2a \cos a} \\ \\ \\ = \frac{\cos (2a-a)}{\sin 2a \cos a} \\ \\ \\ = \frac{\cancel{\cos a}}{\sin 2a \, \cancel{\cos a}} \\ \\ \\ = \frac{1}{\sin 2a} \\ \\ \\ = cosec \, 2a \\ \\ \\ = \mathbb{RHS} \\ \\ \\ \mathbb{H}\mathfrak{ence} \, \, \mathbb{P}\mathfrak{roved}$

Answer 2

To prove:-
$\cot 2 a + \tan a = \cosec 2a$
LHS:-
cot2a + tana
→ (cos2a/sin2a) + tana

→. cos2a/2sina.cosa + sina/cosa

→1/cosa{ (cos2a +2sin²a)/2sina}

→1/2sina.cosa{ 1- 2sin²a + 2sin²a}

→1/sin2a

→cosec2a.
Identities used:-
1. cotø = cosø/sinø
2. sin2ø = 2sinø.cosø
3.cos2ø = 1- 2sin²ø
4. 1/sinø = cosecø.