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Question #a33f6

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Oct 13, 2016

$RHS=sec2theta+tan2theta$

$=(1+tan^2theta)/(1-tan^2theta)+(2tantheta)/(1-tan^2theta)$

$=(1+tantheta)^2/((1-tantheta)(1+tantheta))$

$=(1+tantheta)/(1-tantheta)$

$=(tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta)$

$=tan(pi/4+theta)=LHS$

Proved

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