Trig Problem(s)?

Question

Prove:

sec(x) + tan(y) = cos(y)/(1-siny)

I got this far:

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

Hmm, did I write the problem down right?

Also, if you are feeling especially generous (you don't have too :)) please simplify:

tan(x) - sin(x)cos(x)/sin²(x)

Hope its easy to follow, thank you for your time!

Answer 1

On the first one, they should either be all x's or all y's.
sec x + tan x = cos x / (1 - sin x)
<=> (1/cos x + sin x / cos x) = cos x / (1 - sin x)
<=> (1 + sin x) / cos x = cos x / (1 - sin x)
Multiply through by cos x (1 - sin x):
<=> (1 + sin x) (1 - sin x) = cos^2 x
<=> 1 - sin^2 x = cos^2 x
which is an identity, QED.

(tan x - sin x cosx) / sin^2 x
= (sin x / cos x - sin x cos x) / sin^2 x
= (1 / cos x - cos x) / sin x
= (1 - cos^2 x) / (sin x cos x)
= sin^2 x / (sin x cos x)
= sin x / cos x
= tan x.

Answer 2

Yes, you are on the right track. There's a trick in simplifying trig identities, which it to multiply both numerator and denominator by the conjugate. Multiply (1+sinx) / cosx by (1-sinx)/(1-sinx) to get:

(1-sin²x) / cosx(1-sinx) = cos²x / cosx(1-sinx) = cosx / (1-sinx)

you have mixed up x's and y's, but of course this has to be for the same "argument," x or y.

Answer 3

Assuming you meant sec(y), The LHS is
1/cos(y)+sin(y)/cos(y) = (1+sin(y))/cos(y)
Multiply the RHS by (1+sin(y))/(1+sin(y)).
You get [cos(y)(1+sin(y)]/[1-sin^2(y)] =
[cos(y)(1+sin(y)]/[cos^2(y)] =
(1+sin(y))/cos(y)