Sec thita -1÷ sec thita +1 =(sin thita ÷ 1+ costhita)^2 ?

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Answer 1 · 2018-05-22T09:48:26

Please see the proof below

We need

#sectheta=1/costheta#

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

Therefore the

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

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

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

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

#=(1-cos^2theta)/(1+costheta)^2#

#sin^2theta/(1+costheta)^2#

#=(sintheta/(1+costheta))^2#

#=RHS#

#QED#

Answer 2 · 2018-05-22T10:41:05

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

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

#=(1-cosx)/(1+cosx)*[(1+cosx)/(1+cosx)]#

#=(1-cos^2x)/(1+cosx)^2=sin^2x/(1+cosx)^2=(sinx/(1+cosx))^2=RHS#

Answer 3 · 2018-05-22T10:44:59

Explanation in below

#(secx-1)/(secx+1)#

=#((secx-1)*(secx+1))/(secx+1)^2#

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

=#(tanx)^2/(secx+1)^2#

=#(sinx/cosx)^2/(1/cosx+1)^2#

=#((sinx)^2/(cosx)^2)/((1+cosx)^2/(cosx)^2)#

=#(sinx)^2//(1+cosx)^2#

=#(sinx/(1+cosx))^2#