I proving two Trig identities

[xxiangel]

1. Cos x (sec x + cos x csc^2 x) = csc^2 x

I got as far as this... 1 + cos^2 + cos/sin^2 = csc^2

2. tan x(sin x + cot x cos x) = sec x

 

[moose]

1. Change everything on the left into terms of cos and sin. Then distribute the cosx, after that try to combine anything you can, change anything you can to tanx, etc.

2. Again, change everything you can into sin and cos first, then distribute.

A few of the most important things to keep in mind are, when you are done with simplifying things and whatnot, if something is a fraction, combine the terms. In such trig identities, one of the most used basic definitions is tanx=sinx/cosx

 

[0rthodontist]

I got as far as this... 1 + cos^2 + cos/sin^2 = csc^2

Well, you made a mistake somewhere. Substitute in some random angle and you can see that this is not true.

 

[mrtkawa]

hey you, i got this
cosX(secX+cosXcsc^2X)=csc^2x
just solve the left side
cosX[(1/cosX)+(cosx/sin^2X)]=csc^2X
then multiply ,so...
cosX(1/cosX)+cosX(cosX/sin^2X)=csc^2X
1+cot^2X=csc^2X
since 1+cot^2X one of the trig identity which equals
to csc^2X, problem solved

 

[z-component]

For the future, mrtkawa, have the original poster attempt his/her own work instead of providing the full solution.

 

[konartist]

I was able to solve this till 1+cot^2 = Csc^2 , but do you just use pythagorean identity to fine the identity or what? How are these two equal?

 

[konartist]

for 2.

change everything to cos and sin

SinX/CosX[SinX + CosX/SinX(CosX)] = 1/Cosx

work inside the bracket now.

Cosx/sinx(cosx) = cos^2x/sinx
SinX + Cos^2x/Sinx Now get common denominators
you should notice something and be able to work from there.