Trigonometric Identity

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

Trigonometric Identity

Postby Guest » Thu Mar 14, 2013 4:42 pm

Prove the identity:

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

Postby Guest » Fri Mar 15, 2013 6:26 am

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

If instead you meant
[tex]\frac{\sec^4 y - \tan^4 y}{\sec^2 y + \tan^2 y}[/tex]
then
[tex]\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}[/tex]
[tex]= \sec^2 y - \tan^2 y[/tex]
[tex]= \frac{1}{\cos^2 y} - \frac{\sin^2 y}{\cos^2 y}[/tex]
[tex]= \frac{1 - \sin^2 y}{\cos^2 y}[/tex]
[tex]= \frac{\cos^2 y}{\cos^2 y}[/tex]
[tex]= 1[/tex]

Hope this helped,

R. Baber.
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