Proving tan x + cot x = (sec x)(csc x) and tan² x - sin² x = (tan² x)(sin² x)

[chase222]

I need help solving 2 proofs:

tan x + cot x = (sec x)(csc x)

I changed the left side to:

tan x + 1/tan x = (sec x)(csc x)

then crossed out the tan:

1 = (sec x)(csc x), but I got stuck there.

The next one I had trouble with was:

tan^2 x - sin^2 x = (tan^2 x)(sin^2 x)

I saw the left side being a^2 - b^2, so I factored it into:

(tan x + sin x)(tan x - sin x) = (tan^2 x)(sin^2 x)

I then changed the tan into sin/cos:

((sin x/cos x) + sin x)) ((sin x/cos x) - sin x)) , but got stuck there.

Can you help me solve these proofs?

 

[dextercioby]

Try to write everything in terms of sine & cosine...You'll get them easily.

Daniel.

 

[chase222]

So for the second one:

tan^2 x - sin^2 x = (tan^2 x)(sin^2 x)

(sin^2 x/cos^2 x) - sin^2 X = (sin^2 x/cos^2 x)(sin^2x)

So on both sides so the sin^2 x cancel, leaving it like:

cos^2 x = cos^2 x?

And for the first one:

tan x + cot x = (sec x)(csc x)

I changed it to:

sin x/cos x + 1/(sin x/cos x) = (1/cos x)(1/sin x)

What would I do from here?

 

[dextercioby]

Bring it to the same denominator (in the LHS) and after simplifying the denominators,u'll find

$$\sin^{2}x+\cos^{2}x =1$$



Daniel.

 

[HallsofIvy]

I changed the left side to:

tan x + 1/tan x = (sec x)(csc x)

then crossed out the tan:

1 = (sec x)(csc x), but I got stuck there.

"crossed out the tan" is not a mathematics term! I'm serious- think about exactly what you are doing there. tan x+ 1/tan x is NOT equal to 1 for all x!

(sin^2 x/cos^2 x) - sin^2 X = (sin^2 x/cos^2 x)(sin^2x)

So on both sides so the sin^2 x cancel, leaving it like:

cos^2 x = cos^2 x?

Okay, sin^2 x/cos^2 x) - sin^2 x= (sin^2 x)((1/cos^2x) - 1) so canceling sin^2 x leaves (1/cos^2 x)- 1 = sin^2 x/cos^2 x

That is NOT "cos^2 x= cos^2 x" but if you multiply both sides by cos^2 x you get something almost as easy.