Help!! Trigonometric Identities Question!

Prove! Show the substitutions used

sin x/cot^2 x - sin x/ cos^2 x = -sin x

Hi.

(sinx / cot²x) - (sinx / cos²x)

Substitute cot²x = cos²x/sin²x:
= (sinx / (cos²x/sin²x)) - (sinx/cos²x)
= (sinx)(sin²x/cos²x) - (sinx/cos²x)
= (sin³x/cos²x) - (sinx/cos²x)

Denominator is the same, so combine the sine's on the numerator:
= (sin³x - sinx) / (cos²x)

Factor out a sine in the numrator:
= [sinx(sin² - 1)] / (cos²x)
= [sinx(-1 + sin²x)] / (cos²x)
= [sinx(-1)(1 - sin²x)] / (cos²x)
= [-sinx(1 - sin²x)] / (cos²x)

Substitue 1 - sin²x = cos²x:
= [-sinx(cos²x)] / (cos²x)

Cancel out cos²x from both numerator and denominator:
= (-sinx) / 1
= -sinx

Thus,
(sinx/cot²x) - (sinx/cos²x) = -sinx

Hope this helps :D

sin x/cot^2 x - sin x/cos^2 x = sin x/(cos^2 x/sin^2 x) - sin x/cos^2 x = sin^3 x/cos^2 x - sin x/cos^2 x = (sin^3 x - sin x)/cos^2 x = sin x(sin^2 x - 1)/cos^2 x = sin x(-cos^2 x)/cos^2 x = -sin x

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

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

= sin x [ tan^2(x) - sec^2(x) ]

= sin x [ tan^2(x) - (1 + tan^2(x)) ]

= - sin x