How to start off a proof?

Question

Trouble proving (sinx/1-cotx)-cosx/tanx-1 =sinx+cosx using trig identities

Answer 1

(sinx/[1-cotx])-(cosx/[tanx-1])=sinx+cosx

(sinx/[1-cosx/sinx])-(cosx/[sinx/cosx-1])=sinx+cosx

(sinx/[sinx-cosx/sinx])-(cosx/[sinx-cosx/cosx]=sinx+cosx

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

(sin²x-cos²x)/(sinx-cosx)=sinx+cosx

(sinx-cosx)(sinx+cosx)/(sinx-cosx)=sinx+cosx

sinx+cosx=sinx+cosx

Answer 2

For the denominators you have

 

1 - cot x = 1 - cos x /sin x = (sin x - cos x) / sin x

 

and

 

tan x - 1= sin x/ cos x - 1 = (sin x - cos x) / cos x

 

Hence,

 

sin x / (1 - cot x) - cos x/(tan x - 1)                            (plug in the expressions found above)

= sin^2 x / (sin x - cos x) - cos^2 x / (sin x - cos x)    (combine as the two fractions have the same denominator)

= (sin^2 x - cos^2 x) / (sin x - cos x)                         (use difference of two squares a^2 - b^2 = (a - b)(a + b))

= (sin x - cos x)*(sin x + cos x) / (sin x - cos x)          (cancel)

= sin x + cos x