Need help proving some trig identities

[Confusedbrah]

Proving identities is a pain! Thanks in advance, guys!

Homework Statement

1. 1 + sec^(2)xsin^(2)x = sec^(2)x

2. sinx/1-cosx + sinx/1+cosx = 2cscx

Homework Equations

The Attempt at a Solution

For the first problem, this is the best I got:

1 + sec^(2)x(1-cos(2)x)

For the second problem, I added the fractions together and got:

(sinx + sinxcosx + sinx - sinxcosx) / ((1-cosx) (1-cosx)) =

(2sinx) / (1-cosx) (1+cosx) =

(2sinx) / (1-cos^(2)x) =

(2sinx) / (sin^(2)x)

 

[jedishrfu]

Welcome to PF

For the first one, go back to basics and replace sec with its cos definition ie sec(x)=1/cos(x) and it should be apparent.

For the second one, try combining the fractions on the left and also replace csc(x) with its sin(x) definition.

 

[Confusedbrah]

Welcome to PF

For the first one, go back to basics and replace sec with its cos definition ie sec(x)=1/cos(x) and it should be apparent.

For the second one, try combining the fractions on the left and also replace csc(x) with its sin(x) definition.

Hmmm I kinda see what you're saying. I started all over with the first one and this is what I got, can anyone check if I did it correctly? It proves to be correct, at least to me.

1 + (1/cos^(2)x)(sin^(2))x) = I then multiplied the two fractions, giving me:

1 + (sin^(2)x)/(cos^(2)x) = now I combine the fractions, giving me:

(cos^(2)x + sin^(2)x)/(cos^(2)x) = now I apply a Pythagorean identity to the numerator:

(1)/(cos^(2)x) = sec^(2)x


I am still stumped on the second one. Are you saying combine fractions from the initial equation or the one I worked on and got? Thanks

 

[verty]

Combine the LHS of number 2. Do you see why that is a good idea?

 

[Confusedbrah]

Combine the LHS of number 2. Do you see why that is a good idea?

I think I got it now. For number 2, I separated the two fractions from the left side of the initial equation and got this:

(sinx)/(1) - (sinx)/(cosx) + (sinx)/(1) +(sinx)/(cosx) = then I combined similar terms and got this:

(sinx)/(1) + (sinx)/(1) = now I flipped the fractions and got this:

(1)/(sinx) + (1)/(sinx) = (2)/(sinx) = 2cscx

Can anyone check if I did it correctly? Thanks