Show that (sin theta +cos theta)– (sin theta – cos theta)^2 = 2 sin 2theta.
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Step-by-step explanation:
R.T.P: (sinθ+cosθ)2+(sinθ−cosθ)2=2
LHS:-
(sinθ+cosθ)2+(sinθ−cosθ)2
⟹(sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ−2sinθcosθ)
⟹(sin2θ+cos2θ+sin2θ+cos2θ)
=(1+1)
=2
RHS:-
2
∴ LHS = RHS
Hence, it is proved.
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