Question

(sin^3 theta +cos^3 theta)(sin theta + cos theta )=1-sin theta * cos theta

Answer 1

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correct Question :-

$\frac{(sin^3θ+cos^3 θ)}{(sinθ+ cosθ)}=1-sin θ * cos θ$

step by step solution :-

• let sinθ be x
• let cosθ be y

putting x and y in place of sinθ and cosθ

$\frac{(x^3 + y^3)}{(x+y)}=1-xy$

• $x^3 + y^3 = (x+y)(x^2 -xy + y^2)$

LHS :-

$\frac{(x+y)(x^2 -xy + y^2)}{x+y}$

$(x^2 -xy + y^2)$

RHS:-

$1 - xy$

compare:-

$(x^2 -xy + y^2) = 1 - xy$

• putting sinθ and cosθ in place of x and y

$(sinθ^2 -sinθ*cosθ+ cosθ^2) = 1 - sinθ*cosθ$

• $sinθ^2 + cosθ^2 = 1$

$(1 -sinθ*cosθ) = 1 - sinθ*cosθ$

LHS. = RHS

..............hence verified

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