How to prove the trigonometric identity?

Question

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

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Answer 1 · 2018-06-04T11:00:38

Answer for the Question:

#(1+sintheta-costheta)/(1+sintheta+color(red)(costheta))+(1+sintheta+costheta)/(1+sintheta-color(red)(costheta))=2cosectheta#

Here,

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

We take,

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

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

#={(1+sin^2theta+cos^2theta+2sintheta-2sinthetacostheta-2costheta+#

#color(white)(.....)((1+sin^2theta+cos^2theta+2sintheta+2sinthetacostheta+2costheta}) /((1+sintheta)^2-(costheta)^2#

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

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

#=(2(2+2sintheta))/(2sintheta+2sin^2theta)#

#=(2(2+2sintheta))/(sintheta(2+2sintheta))#

#=2/sintheta#

#=2csctheta#

#=RHS#