Plzz solve the 6th question!!
The ques is in the attachment below.
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HEYA MATE, HERE IS UR ANSWER ________________
Taking LHS
sinA-sin^3A/2cos^3A-cosA
=sinA (1-sin^2A)/cosA (2cos^2A-1)
we know that sinA/Cos A=tanA
=TanA {(1-2 (1-cos^2A)/2cos^2A-1}(as sin^2A=1-cos^2A)
=TanA {1-2+2cos^2A/2cos^2A-1}
=TanA {-1+2cos^2A/2cos^2A-1}
=TanA {2cos^2A-1/2cos^2A-1}
=TanA
=RHS .
________☆☆HOPE THE ANSWER HELPS YOU ☆☆____☆☆BE BRAINLY ☆☆___☆☆
Taking LHS
sinA-sin^3A/2cos^3A-cosA
=sinA (1-sin^2A)/cosA (2cos^2A-1)
we know that sinA/Cos A=tanA
=TanA {(1-2 (1-cos^2A)/2cos^2A-1}(as sin^2A=1-cos^2A)
=TanA {1-2+2cos^2A/2cos^2A-1}
=TanA {-1+2cos^2A/2cos^2A-1}
=TanA {2cos^2A-1/2cos^2A-1}
=TanA
=RHS .
________☆☆HOPE THE ANSWER HELPS YOU ☆☆____☆☆BE BRAINLY ☆☆___☆☆
SillySam:
my pleasure sis :)
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