I NEED help for my TRIG homework.?

Question

Please and thank you.

PROVE THAT

[1 - cosx over sinx] + [ sinx over 1- cosx] = 2 cosecx

Answer 1

(1 - cos x) / sin x + sin x / (1 - cos x)
= [(1 - cos x)^2 + sin^2 x] / [sin x (1 - cos x)]
= [1 - 2 cos x + cos^2 x + sin^2 x] / [sin x (1 - cos x)]
= [2 - 2 cos x] / [sin x (1 - cos x)]
= [2 (1 - cos x)] / [sin x (1 - cos x)]
= 2 / sin x
= 2 cosec x.

Answer 2

[(1 - cos x) / sin x] +[sin x / (1 - cos x)] = 2 cosec x

simplify left by getting least common denominator (lcd) which is (sin x)(1-cos x)

divide lcd by the denominator of the first term gives you (1 - cos x) (1 - cos x) and divide lcd by denominator of the the second term gives you (sin x) (sin x)

simplify further, left term becomes (1 - 2cos x + cos^2x + sin^2)

since sin^2 + cos^2 = 1, the numerator becomes 1 - 2cosx + 1 = 2 - 2cosx

factoring out 2 gives: 2(1 - cosx)

the new fraction now is [2(1 - cos x) / (1 - cos x)(sin x)]

cancelling (1 - cos x) gives us 2 / sinx

since 1/sin x = cosec x

final answer: 2 cosec x