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 :)

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