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

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

Ans:

Let’s start from LHS

LHS = $\frac{\cos^2 \theta}{1-\tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}$

= $\frac{\cos^2 \theta}{1-\frac{\sin \theta}{\cos\theta}} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}$

= $\frac{\cos^2 \theta}{\frac{\cos\theta-\sin \theta}{\cos\theta}} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta}$

= $\frac{\cos^3 \theta}{\cos\theta-\sin \theta} - \frac{\sin^3 \theta}{\cos \theta - \sin \theta}$

= $\frac{\cos^3 \theta - \sin^3\theta}{\cos\theta-\sin \theta}$

We know that, a³ – b³ = (a – b) (a² + b² + ab)

$\therefore$ LHS = $\frac{(\cos \theta - \sin\theta)(\cos^2\theta + \sin^2\theta + \cos \theta \sin \theta)}{\cos \theta - \sin \theta}$

= (cos² θ + sin²θ + sin θ cos θ)

= 1 + sinθ cosθ

= RHS …………. Hence proved!

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