Cos^2theta /1-tan theta+sin^2theta/1-cot theta=1+sin theta cos theta

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Solution

$LHS = \frac{\cos^{2} θ}{1 - \tan θ} + \frac{\sin^{2} θ}{1 - \cot θ} = \frac{\cos^{2} θ}{1 - \frac{\sin θ}{\cos θ}} + \frac{\sin^{2} θ}{1 - \frac{\cos θ}{\sin θ}} = \frac{\cos^{3} θ}{(\cos θ - \sin θ)} + \frac{\sin^{3} θ}{(\sin θ - \cos θ)} = \frac{\cos^{3} θ}{(\cos θ - \sin θ)} - \frac{\sin^{3} θ}{(\cos θ - \sin θ)} = \frac{\cos^{3} θ - \sin^{3} θ}{(\cos θ - \sin θ)} = \frac{(\cos θ - \sin θ) (\cos^{2} θ + \sin^{2} θ + \cos θ . \sin θ)}{(\cos θ - \sin θ)} = (\cos^{2} θ + \sin^{2} θ + \cos θ . \sin θ) = 1 + \sin θ . \cos θ = RHS$

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