Question #a32b3

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Answer 1 · 2017-03-13T19:36:22

See proof below

We use

$(a+b)^2=a^2+2ab+b^2$

$sin^2x+cos^2x=1$

Dividing by $cos^2x$

$sin^2x/cos^2x+1=1/cos^2x$

$tan^2x+1=sec^2x$

$tan^2x=sec^2x-1$

Therefore,

$LHS=(tanx+secx)^2$

$=tan^2x+sec^2x+2tanxcosx$

$=sec^2x-1+sec^2x+2tanxsecx$

$=2sec^2x+2tanxsecx-1$

$=RHS$

$QED$