Question #2cf23

2 Answers
Nov 25, 2017

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

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

#=(sec^2A-1)/(secA+1)^2#

#=tan^2A/(1/cosA+1)^2#

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

#=(sinA/(1+cosA))^2=RHS#

Nov 25, 2017

See the proof below

Explanation:

We need

#secA=1/cosA#

#cos^2A+sin^2A=1#

Therefore,

#LHS=(secA-1)/(secA+1)=(1/cosA-1)/(1/cosA+1)#

#=(1-cosA)/(1+cosA)#

#=((1-cosA)(1+cosA))/((1+cosA)(1+cosA))#

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

#=(sin^2A)/(1+cosA)^2#

#=((sinA)/(1+cosA))^2#

#=RHS#

#QED#