Prove the trigonometric identity ?

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

2 Answers
Sonnhard
Jun 2, 2018

It is equivalent to
#sin^2(x)+cos^2(x)=1#

Explanation:

By cross multiplication we get
#sin(x)(sin(x)-1-cos(x))=#
#sin^2(x)-sin(x)-sin(x)cos(x)#
The Right-Hand side is given by
#(1+cos(x))(1-sin(x)-cos(x))=#
#1-sin(x)-cos(x)+cos(x)-sin(x)cos(x)-cos^2(x)#
the equal sign holds if
#sin^2(x)+cos^2(x)=1#

P dilip_k
Jun 2, 2018

#RHS=(1-sintheta-costheta)/(sintheta-1-costheta)#

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

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

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

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

#=sintheta/(1+costheta)=LHS#

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