Prove that: cos^2 θ/(1 – tan θ) + sin^3 θ/(sin θ – cos θ) = 1 + sin θ cos θ

Question

Q) Prove that: $\frac{\cos^2 	heta}{1-	an 	heta} +  \frac{\sin^3 	heta}{\sin 	heta - \cos 	heta}$ =  1 + sinθ cosθ

Sapling Academy · Accepted Answer

Let’s start from LHS
LHS = $\frac{\cos^2 	heta}{1-	an 	heta} +  \frac{\sin^3 	heta}{\sin 	heta - \cos 	heta}$
= $\frac{\cos^2 	heta}{1-\frac{\sin 	heta}{\cos	heta}} +  \frac{\sin^3 	heta}{\sin 	heta - \cos 	heta}$
= $\frac{\cos^2 	heta}{\frac{\cos	heta-\sin 	heta}{\cos	heta}} +  \frac{\sin^3 	heta}{\sin 	heta - \cos 	heta}$
= $\frac{\cos^3 	heta}{\cos	heta-\sin 	heta} -  \frac{\sin^3 	heta}{\cos 	heta - \sin 	heta}$
= $\frac{\cos^3 	heta - \sin^3	heta}{\cos	heta-\sin 	heta}$
We know that, a3 - b3 = (a - b) (a2 + b2 + ab)
$	herefore$ LHS = $\frac{(\cos 	heta - \sin	heta)(\cos^2	heta + \sin^2	heta + \cos 	heta \sin 	heta)}{\cos 	heta - \sin 	heta}$
= (cos2  θ + sin2 θ + sin θ cos θ)
= 1 + sinθ cosθ
= RHS ............. Hence proved!

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