Sin theta × cos( 90 -theta) + cos theta × sin (90 - theta) = 1
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sin θx cos(90- θ)+cos θx sin (90- θ) =1
sin θ x sinθ + cosθ x cos θ. {because cos(90- θ)=sin θ} and similarly with sin θ
sin² θ+cos² θ=1
LHS=RHS
sin θ x sinθ + cosθ x cos θ. {because cos(90- θ)=sin θ} and similarly with sin θ
sin² θ+cos² θ=1
LHS=RHS
Answered by
2
L.H.S = sin θ cos (90° - θ) + cos θ sin (90° - θ) = sin θ sin θ + cos θ cos θ = sin 2 θ + cos 2 θ
[Using Identity sin 2 φ + cos 2 φ = 1]
= 1
= R.H.S.
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