Solve this problem number 34
It's from trigonometrical identities
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Taking LHS
SinA-2 sin^3A /2cos^3A-cosA
= SinA (1-2sin^2A)/cosA (2cos^2A-1 )
=TanA { 1-2(1-cos^2A )}/(2cos^2A-1) _________{SinA/cosA=TanA and Sin^2A=1-cos^2A}
=TanA { 1-2+2cos^2A}/(2cos^2A-1)
=TanA { 2cos^2A-1}/(2cos^2A-1)
=TanA
=RHS
__________Hence verified _____
______☆☆HOPE THE ANSWER HELPS YOU. ☆☆_______☆☆BE BRAINLY ☆☆☆
Taking LHS
SinA-2 sin^3A /2cos^3A-cosA
= SinA (1-2sin^2A)/cosA (2cos^2A-1 )
=TanA { 1-2(1-cos^2A )}/(2cos^2A-1) _________{SinA/cosA=TanA and Sin^2A=1-cos^2A}
=TanA { 1-2+2cos^2A}/(2cos^2A-1)
=TanA { 2cos^2A-1}/(2cos^2A-1)
=TanA
=RHS
__________Hence verified _____
______☆☆HOPE THE ANSWER HELPS YOU. ☆☆_______☆☆BE BRAINLY ☆☆☆
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