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HELLO DEAR,
GIVEN:-
we know:- sin2θ = 2sinθcos
cos2θ = 2cos²θ - 1
now,
sin2θ/(1 + cos2θ)
=> 2sinθcosθ/{1 + 2cos²θ - 1}
=> 2sinθcosθ/2cos²θ
=> sinθ/cosθ
=> tanθ
I HOPE IT'S HELP YOU DEAR,
THANKS
GIVEN:-
we know:- sin2θ = 2sinθcos
cos2θ = 2cos²θ - 1
now,
sin2θ/(1 + cos2θ)
=> 2sinθcosθ/{1 + 2cos²θ - 1}
=> 2sinθcosθ/2cos²θ
=> sinθ/cosθ
=> tanθ
I HOPE IT'S HELP YOU DEAR,
THANKS
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