How do you prove $sec2theta=sec^2theta/(2-sec^2theta)$ ?

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Answer 1 · 2016-08-19T22:24:25

$"To prove: "sec2theta=sec^2theta/(2-sec^2theta)$

$RHS=sec^2theta/(2-sec^2theta)$

$=(cos^2thetasec^2theta)/(cos^2theta(2-sec^2theta))$

$=1/(2cos^2theta-cos^2thetasec^2theta)$

$=1/(2cos^2theta-1)$

$=1/cos(2theta)=sec2theta=LHS$

PROVED

How do you prove sec2theta=sec^2theta/(2-sec^2theta)sec2theta=sec^2theta/(2-sec^2theta)?