Math, Verify

Cosx + SinxTanx = SecX

I have been looking at this for some time now and have no idea. I am more than likely doing something supid, but I just can't figure it out. Pre-Calculus by the way. 

This is 10th grade shit.

Like I already stated, Pre-Calculus.

Express Tan and Sec using Sin and Cos and you should be able to see the relationship.

Jesus Christ, dude. If you're ever trying to prove a trig identity, the first thing you do always is make everything sine and cosine.

lol people still use sec and not 1/cos

multiply the terms on the left side by cos/cos, simplify to (cos^2 + sin^2)/cos, use pythagorean identity to re-express as 1/cos, and now it equals the right side.

cos(x) + sin(x)tan(x) = sec(x)
tan(x) = sin(x)/cos(x)
sec(x) = 1/cos(x)
cos(x) + sin²(x)/cos(x) = 1/cos(x)
cos²(x)/cos(x) + sin²(x)/cos(x) = 1/cos(x)
cos²(x) + sin²(x) = 1
nigger.

>>8
He said verify not solve idiot.


cos(x)+sin(x)tan(x) = sec (x)
(sinx)(sinx/cosx) + cos x
sin^2 x/cosx) + cos x
(sin^2 x + cos^2 x)/cos x
1/cos x
sec x

>>9
You fucking moron.

>>10
Seconded.

>>9
Hurf fucking durf.

You're going to love this...


LHS:

CosX + SinXTanX         (Keep TanX as TanX, it makes moveing around the numbers easier)

Divide both terms by CosX:

1 + (SinXTanX)/(CosX)

Divide both terms by SinXTanX:

(1)/(SinXTanX) + (1)/(CosX)

Simplify out TanX:

(CosX)/((SinX)^2) + (1)/(CosX)

Substitute (SinX^2) for Pythagorean Identity ((SinX)^2) = 1 - (CosX)^2) and cancel CosX:

(1)/(1-CosX) + (1)/(CosX)

Divide both terms by 1 - CosX:

1 + (1 - CosX)/(CosX)

Separate out fractions:

1 + (1)/(CosX) - (CosX)/(CosX)

Simplify CosX:

1 + (1)/(CosX) - 1

1-1 = 0, hence LHS = RHS

QED.