prove that sin theta +sin2 theta/1+cos theta+cos2 theta=tan theta
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Answered by
101
use identities ,
sin2∅ = 2sin∅.cos∅
cos2∅ = 2cos²∅ -1
( sin∅ + sin2∅)/(1 + cos∅ + cos2∅)
=( sin∅ + 2sin∅.cos∅)/(1 + cos∅ +2cos²∅ -1)
=sin∅( 1+ 2cos∅)/cos∅( 1 +2cos∅)
=tan∅ = RHS
sin2∅ = 2sin∅.cos∅
cos2∅ = 2cos²∅ -1
( sin∅ + sin2∅)/(1 + cos∅ + cos2∅)
=( sin∅ + 2sin∅.cos∅)/(1 + cos∅ +2cos²∅ -1)
=sin∅( 1+ 2cos∅)/cos∅( 1 +2cos∅)
=tan∅ = RHS
Answered by
27
Answer:
Consider a right-angled triangle, ΔABC, where ∠BAC=θ,

( sin∅ + sin2∅)/(1 + cos∅ + cos2∅)
=( sintheta + 2sintheta.costheta)/(1 + costheta +2cos²theta -1)
=sintheta( 1+ 2costheta)/costheta( 1 +2costheta)
=tantheta = RHS
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