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

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

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

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

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

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

$=sec^2thetacsc^2theta$