Prove that, #(1+sintheta-costheta)^2/(1+sintheta+costheta)^2=(1-costheta)/(1+costheta)# ?

2 Answers
maganbhai P.
Jun 11, 2018

Please see below.

Explanation:

We know that,
#color(red)((1)(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca#

#color(blue)((2)(a+b-c)^2=a^2+b^2+c^2+2ab- 2bc-2ca#

We have to prove,

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

We take,

#LHS=(1+sintheta-costheta)^2/(1+sintheta+costheta)^2...toApply (1) and (2)#

#color(white)(LHS)=(1+color(brown)(sin^2theta+cos^2theta)+2sintheta- 2costheta- sinthetacostheta)/(1+color(brown)(sin^2theta+cos^2theta)+2sintheta+2costheta +sinthetacostheta)#

#color(white)(LHS)=(1+1+2sintheta-2costheta-sinthetacostheta)/ (1+1+2sintheta+2costheta+sinthetacostheta)#

#color(white)(LHS)=(2+2sintheta-2costheta-sinthetacostheta)/ (2+2sintheta+2costheta+sinthetacostheta)#

#color(white)(LHS)= (2(1+sintheta)-2costheta(1+sintheta))/(2(1+sintheta)+2costheta(1 +sintheta)#

#color(white)(LHS)=(cancel((1+sintheta))(2-2costheta))/(cancel((1+sintheta)) (2+2costheta)#

#color(white)(LHS)=(2-2costheta)/(2+2costheta)#

#color(white)(LHS)=(cancel2(1-costheta))/(cancel2(1+costheta))#

#color(white)(LHS)=(1-costheta)/(1+costheta)#

#LHS=RHS#

P dilip_k
Jun 12, 2018

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

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

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

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

#=(2(1-costheta))/(2(1+costheta))#

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

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