# Math Help - Somewhat Advanced Trig Identity Problem...

1. ## Somewhat Advanced Trig Identity Problem...

Show: $\frac{1-3\sin x - 4\sin^2 x}{\cos^2 x} = \frac{1 - 4\sin x}{1 - \sin x}$

Your help is greatly appreciated. Thanks.

2. Originally Posted by Savior_Self
Show: $\frac{1-3\sin x - 4\sin^2 x}{\cos^2 x} = \frac{1 - 4\sin x}{1 - \sin x}$

Your help is greatly appreciated. Thanks.
Hi Savior_Self,

Take the right side and multiply by $\frac{1+\sin x}{1+\sin x}$:

$\frac{1-4\sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}=\frac{1-3 \sin x-4 \sin^2 x}{1-\sin^2 x}=\frac{1-3 \sin x - 4 \sin^2 x}{\cos^2 x}$

3. or try factoring the Left Hand Side (LHS) of the equation?

it is a simple factoring:

LHS = (1 - 4sin x)(1 + sin x) =
(1 + sin x)(1 - sin x)

the (1 + sin x) cancels, you are left with,

(1 - 4sin x)/(1 - sin x).

(1 - 4sin 4x)(1 + sin x) = RHS