How do you prove $1/(1+sin(theta)) + 1/(1-sin(theta)) = 2sec^2(theta)$ ?

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Answer 1 · 2016-04-17T00:31:48

See below

LHS =left hand side, RHS =right hand side

LHS $=1/(1+sin theta)+1/(1-sin theta)$

$=(1-sin theta +1+sin theta )/((1+sin theta)(1-sin theta))$ ->Common Denominator

$=(1-cancelsin theta +1+cancelsin theta )/((1+sin theta)(1-sin theta))$

$=2/(1-sin^2x)$

$=2/cos^2x$

$=2*1/cos^2x$

$=2sec^2x$

$=RHS$

How do you prove 1/(1+sin(theta)) + 1/(1-sin(theta)) = 2sec^2(theta)1/(1+sin(theta)) + 1/(1-sin(theta)) = 2sec^2(theta)?