Question #ac0e3

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Answer 1 · 2017-03-22T16:30:34

see explanation

let we take LHS to prove RHS.

$tan 2theta = sin(2theta) / cos (2theta)$

$sin(2theta) = 2sin theta cos theta, and cos (2theta)= cos^2 theta - sin^2 theta$

therefore,
$tan 2theta= (2sin theta cos theta)/ (cos^2 theta - sin^2 theta)$

divide by $cos^2 theta$

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

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

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

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

Answer 2 · 2017-03-22T16:37:25

$LHS=tan2theta$

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

$=(2sinthetacostheta)/(cos^2theta-sin^2theta)$

$=((2sinthetacostheta)/cos^2theta)/((cos^2theta-sin^2theta)/cos^2theta)$

$=((2sintheta)/costheta)/(cos^2theta/cos^2theta-sin^2theta/cos^2theta)$

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