Question

prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx

Answer 1

Answer:

(2sinxcosx-cosx)/(1 -sinx+sin²x-cos²x)=cotx

Step-by-step explanation:

Prove that 2sinxcosx-cosx/1 -sinx+sin^2x-cos^2x=cotx

LHS

= (2SinxCosx - Cosx)/(1 - Sinx + Sin²x - Cos²x)

= (Cosx(2Sinx - 1)/( 1  - Cos²x + Sin²x - Sinx)

Using 1 - Cos²x = Sin²x

= (Cosx(2Sinx - 1)/( Sin²x + Sin²x - Sinx)

= (Cosx(2Sinx - 1)/(2Sin²x  - Sinx)

= (Cosx(2Sinx - 1)/(Sinx(2Sinx  - 1)

Cancelling 2Sinx - 1

= Cosx/Sinx

= Cotx

= RHS

QED

Proved

(2sinxcosx-cosx)/(1 -sinx+sin²x-cos²x)=cotx

Answer 2

Answer:

LHS = (2sinxcosx-cosx)/(1-sinx-cos^2x+sin^2x)

= (2sinxcosx-cosx)/(sin^2x+cos^2x-sinx-cos^2x+sin^2x)

= (2sinxcosx-cosx)/(2sin^2x+cos^2x)

= cosx(2sinx-1)/sinx(2sinx-1)

= cosx/sinx

= cotx

=RHS