Prove that #(1+sin2A)/(1-sin2A)=tan^2(pi/4+A)#?

Question

15417 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-14T10:46:02

#LHS=(1+sin2A)/(1-sin2A)#

#=(cos^2A+sin^2A+2sin2cosA)/(cos^2A+sin^2A-2sinAcosA)#

#=(cosA+sinA)^2/(cosA-sinA)^2#

#=(cosA/cosA+sinA/cosA)^2/(cosA/cosA-sinA/cosA)^2#

#=(1+tanA)^2/(1-tanA)^2#

#=(tan(pi/4)+tanA)^2/(1-tan(pi/4)tanA)^2#

#=tan^2(pi/4+A)=RHS#

Answer 2 · 2017-11-14T10:58:15

Please refer to a Proof in the Explanation.

Sub.ing #pi/4+A=theta," so that, "2A=2(theta-pi/4)=2theta-pi/2.#

#:. (1+sin2A)/(1-sin2A),#

#={1+sin(2theta-pi/2)}/{1-sin(2theta-pi/2)},#

#={1-sin(pi/2-2theta)}/{1+sin(pi/2-2theta)},#

#=(1-cos2theta)/(1+cos2theta),#

#=tan^2theta,#

#=tan^2(pi/4+A).#

Hence, the Proof.

Enjoy Maths!