How does #sec^2(pi/2-theta)+csc^2(pi/2-theta)=sec^2thetacsc^2theta#?

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Show: #sec^2(pi/2-theta)+csc^2(pi/2-theta)=sec^2thetacsc^2theta#

This is the first part of the answer that my answer book gives me.
#sec^2(pi/2-theta)+csc^2(pi/2-theta)=sec^2thetacsc^2theta#
I don't see how #sec^2(pi/2-theta)+csc^2(pi/2-theta)# is equal to #sec^2thetacsc^2theta#.

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Answer 1 · 2017-08-02T01:24:12

#LHS=sec^2(pi/2-theta)+csc^2(pi/2-theta)#

#=csc^2(theta)+sec^2(theta)#

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

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

#=1 /(sin^2(theta)cos^2(theta))=RHS#

#=sec^2thetacsc^2theta#