Simplify #(1+cos2A+sin2A)/(1-cos2A+sin2A)#?

Question

9189 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-18T07:29:22

Please see below.

#(1+cos2A+sin2A)/(1-cos2A+sin2A)#

Now #sin2A=2sinAcosA# and #cos2A=2cos^2A-1=1-2sin^2A#

Hence #(1+cos2A+sin2A)/(1-cos2A+sin2A)#

= #(1+2cos^2A-1+2sinAcosA)/(1-(1-2sin^2A)+2sinAcosA)#

= #(1+2cos^2A-1+2sinAcosA)/(1-1+2sin^2A+2sinAcosA)#

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

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

= #(2cosA)/(2sinA)#

= #cotA#