Question #31b2f

3 Answers
Jim H
Aug 10, 2017

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)#

Subhrajit M.
Aug 11, 2017

Answer is below

Explanation:

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

Ratnaker Mehta
Aug 11, 2017

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!

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