Some proofs :(?

Question

Ugh, I'm really horrible at these. Burning through some homework here and I can't for the life of me figure out
(1+cosx)(cscx - cotx) = sinx

I'm not sure how these ones that involve so much multipling, or really what identities to use. if anyone could help explain this to me, I'll surely thank you with a best answer.

Also, is sin^2x/sin^4x = 1/sin^2x ?

Answer 1

(1 + cos x) (csc x - cot x)

First, multiply out the parentheses:

csc x - cot x + cos x csc x - cos x cot x

Now, convert everything to sines and cosines:

1/sin x - cos x/sin x + cos x/sin x - cos² x/sin x

Make the obvious cancellation:

1/sin x - cos² x/sin x

Now, use the Pythagorean theorem:

1/sin x - (1 - sin² x)/sin x

Expand:

1/sin x - 1/sin x + sin² x/sin x

Make the obvious cancellations:

sin x

And we are done.

Answer 2

You're not going to get much better if you keep thinking "I'm really horrible at these". But anyway:

Multiply the left side out, the same way as you would multiply out something like (a+b)(c-d).. Then make use of the fact that csc(x) = 1 / sin(x), and cot(x) = cos(x)/sin(x)

csc(x) + cos(x)csc(x) - cot(x) - cos(x)cot(x)
csc(x) + cos(x)/sin(x) - cos(x)/sin(x) - cos^2(x)/sin(x)
csc(x) - cos^2(x)/sin(x)

Factor out a 1/sin(x), and use the fact that sin^2(x) + cos^2(x) = 1

(1 - cos^2(x)) / sin(x)
(sin^2(x)) / sin(x)
sin(x)


And yes, sin^2(x)/sin^4(x) = 1/sin^2(x). It's like any other exponent.