Verifying a trig identity

[TheNascar92]

Im suppsed to verify an identity but I am stuck after trying several different ways to solve it...maybe im making a mistake somewhere..the identity is as follows:

secx=cosx+secx(sin^2x)

the last part reads "sine squared x" ive tried using the pythagorean identity nothing is working any help appreciated thanks

 

[whitecanvas]

One way to prove an identity is to convert it to sines and cosines and solve.

Work on the RHS.

First change the sec x to 1/cosx.

You're right, for sin[sup:wl8ydonc]2[/sup:wl8ydonc]x, you should use the Pythagorean Identity:

sin[sup:wl8ydonc]2[/sup:wl8ydonc]x = 1 - cos[sup:wl8ydonc]2[/sup:wl8ydonc]x

So you should have the following equation after converting it to cosines:

cos x + (1/cosx)(1 - cos[sup:wl8ydonc]2[/sup:wl8ydonc]x) =

Now solve it.

Correct me if I'm wrong, because I'm just starting to work on this, too.

 

[soroban]

Hello, TheNascar92!

Whitecanvas had good advice . . . Here's my approach.

\(\displaystyle \text{Verify: }\;\sec x\:=\:\cos x+\sec x\~!\cdot\!\sin^2\!x\)

\(\displaystyle \text{The right side is: }\;\cos x + \frac{1}{\cos x}\!\cdot\!\sin^2\!x \;=\;\cos x + \frac{\sin^2\!x}{\cos x}\)


\(\displaystyle \text{Since }\sin^2\!x \:=\:1-\cos^2\!x,\)

. . \(\displaystyle \text{we have: }\;\cos x + \frac{1-\cos^2\!x}{\cos x}\)

. . . . . . .\(\displaystyle =\;\;\cos x + \frac{1}{\cos x} - \frac{\cos^2\!x}{\cos x}\)

. . . . . . .\(\displaystyle = \;\;\cos x + \frac{1}{\cos x} - \cos x\)

. . . . . . .\(\displaystyle =\;\;\frac{1}{\cos x}\)

. . . . . . .\(\displaystyle =\;\;\sec x \quad\hdots\quad \text{ta-}DAA!\)