sinA-2sin^3A/2cos^3A-cosA=tanA explain,step by step and using statements
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Answered by
17
by using identity sin^2A+cos^2A.
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Answered by
11
Hi ,
LHS = (sinA - 2sin³A)/(2cos³A-cosA)
= [SinA(1 - 2sin²A)]/[cosA (2cos²A-1)]
= tanA{[1 - 2(1-cos²A)]}/(2cos²A - 1 )
= TanA [ 1 - 2+2cos²A ]/(2cos²A - 1 )
= TanA [ 2cos²A - 1 ]/ ( 2cos²A - 1 )
= tanA
= RHS
I hope this helps you.
: )
LHS = (sinA - 2sin³A)/(2cos³A-cosA)
= [SinA(1 - 2sin²A)]/[cosA (2cos²A-1)]
= tanA{[1 - 2(1-cos²A)]}/(2cos²A - 1 )
= TanA [ 1 - 2+2cos²A ]/(2cos²A - 1 )
= TanA [ 2cos²A - 1 ]/ ( 2cos²A - 1 )
= tanA
= RHS
I hope this helps you.
: )
sanyogitad:
thanks
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