Question

Question #31b2f

Answer 1

Please see below.

Explanation:

`(1+tanx)/(1-tanx) = (1+(sinx/cosx))/(1-(sinx/cosx))`

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

`= ((2sinx))/((2sinx)) * ((cosx+sinx))/((cosx-sinx))`

`= (2sinxcosx+2sin^2x)/(2sinxcosx-2sin^2x)`

`= (sin2x+1-1+2sin^2x)/(sin2x -1+1-2sin^2x)`

`= (sin2x+1-(1-2sin^2x))/(sin2x -1+(1-2sin^2x))`

`= (sin2x+1-cos2x)/(sin2x -1+cos2x)`

Answer 2

Answer is below

Explanation:

![https://useruploads.socratic.org/nTwt7AcSTii8k1VoQILa_IMG-20170811-WA0009.jpg)

Answer 3

Refer to a Proof in the Explanation.

Explanation:

Knowing that,

`sin2x=2sinxcosx, and, 1-cos2x=2sin^2x,` we have,

`(sin2x)/(1-cos2x)=(2sinxcosx)/(2sin^2x)=cosx/sinx=cotx,`

`:. (sin2x)/(1-cos2x)=1/tanx.`

By, Componendo-Dividendo, then, we get,

`(sin2x+1-cos2x)/(sin2x-(1-cos2x))=(1+tanx)/(1-tanx).`

Hence, the Proof.

Enjoy Maths!