# Trigonometric Identity

Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot

### Trigonometric Identity

Prove the identity:

(sec4y-tan2y)/(sec2y+tan2y)=1
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### Re: Trigonometric Identity

$$\frac{\sec^4 y - \tan^2 y}{\sec^2 y + \tan^2 y} = 1+\frac{\sec^4 y - \tan^2 y -\sec^2 y - \tan^2 y}{\sec^2 y + \tan^2 y}$$
$$= 1+\frac{1 - 2\sin^2 y \cos^2 y -\cos^2 y}{\cos^2 y + \sin^2 y \cos^2 y}$$
$$= 1+\frac{sin^2 y - 2\sin^2 y \cos^2 y}{\cos^2 y + \sin^2 y \cos^2 y}$$
$$= 1+\frac{sin^2 y(1 - 2 \cos^2 y)}{\cos^2 y + \sin^2 y \cos^2 y}$$
$$= 1-\frac{sin^2 y \cos 2y}{\cos^2 y + \sin^2 y \cos^2 y}$$
which equals $$1$$ if and only if $$sin^2 y \cos 2y = 0$$,
i.e. $$y = \pi n$$, $$-\frac{\pi}{4} + \pi n$$, or $$\frac{\pi}{4} +\pi n$$ for some integer $$n$$.

$$\frac{\sec^4 y - \tan^4 y}{\sec^2 y + \tan^2 y}$$
then
$$\frac{\sec^4 y - \tan^4 y}{\sec^2 y + \tan^2 y} = \frac{(\sec^2 y - \tan^2 y)(\sec^2 y + \tan^2 y)}{\sec^2 y + \tan^2 y}$$
$$= \sec^2 y - \tan^2 y$$
$$= \frac{1}{\cos^2 y} - \frac{\sin^2 y}{\cos^2 y}$$
$$= \frac{1 - \sin^2 y}{\cos^2 y}$$
$$= \frac{\cos^2 y}{\cos^2 y}$$
$$= 1$$

Hope this helped,

R. Baber.
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