How to prove #(sin(theta))/(1+sin(theta))=1-[(1-sin(theta))/(cos^2(theta))]?#

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Answer 1 · 2018-03-01T04:15:04

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

#=(cos^2x-(1-sinx))/cos^2x#

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

#=(cancel((1-sinx))[1+sinx-1])/(cancel((1-sinx))(1+sinx))#

#=sinx/(1+sinx)=LHS#

Answer 2 · 2018-03-01T07:21:51

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

#LHS=(sintheta*cos^2theta+ (1-sintheta)(1+sintheta)) /((1+sintheta) cos^2theta)#

#=(sintheta*cos^2theta+ (1-sin^2theta)) /((1+sintheta) cos^2theta)#

#=(sintheta*cos^2theta+ cos^2theta) /((1+sintheta) cos^2theta)#

#=(cancel((sintheta+ 1))cancelcos^2theta) /(cancel((1+sintheta)) cancelcos^2theta)#

#=1 = RHS#

Hence Proved.

-Sahar :)