Question

Select the correct answer from each drop-down menu.

Answer 1

Answer:

$(b)\\\tan x, \cot x$

Step-by-step explanation:

Given

trigonometric function is

$\dfrac{\cos 2x-\cos 4x}{\sin 2x+\sin 4x}$

Applying trigonometric formulae

$\Rightarrow \cos C-\cos D=-\sin (\dfrac{C+D}{2})\sin (\dfrac{C-D}{2})\\\\\Rightarrow \sin C+\sin D=\sin (\dfrac{C+D}{2})\cos (\dfrac{C-D}{2})$

$\Rightarrow \dfrac{\cos 2x-\cos 4x}{\sin 2x+\sin 4x}=\dfrac{2\sin 3x\cdot \sin x}{2\sin 3x\cdot \cos x}\\\\\Rightarrow \dfrac{\sin x}{\cos x}=\tan x$

option (b) is correct

$\Rightarrow \dfrac{\cos 2x+\cos 4x}{\sin 2x-\sin 4x}=\dfrac{2\cos 3x\cdot \cos x}{2\cos 3x\cdot \sin (-x)}=-\cot x$