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Question: \(1 - \frac{\sin^{3}\theta}{\sin\theta + \cos\theta} - \frac{\cos^{3}\theta}{\sin\theta + \cos\theta...

1−sin⁡3θsin⁡θ+cos⁡θ−cos⁡3θsin⁡θ+cos⁡θ1 - \frac{\sin^{3}\theta}{\sin\theta + \cos\theta} - \frac{\cos^{3}\theta}{\sin\theta + \cos\theta}is equal to

A

sin⁡2θ\sin^{2}\theta

B

cos⁡2θ\cos^{2}\theta

C

sin⁡θcos⁡θ\sin\theta\cos\theta

D

sin⁡2θ\sin^{2}\theta

Answer

sin⁡θcos⁡θ\sin\theta\cos\theta

Explanation

Solution

We can write the given expression as

1−sin⁡3θ+cos⁡3θsin⁡θ+cos⁡θ=1−(sin⁡θ+cos⁡θ)(sin⁡2θ+cos⁡2θ−sin⁡θcos⁡θ)sin⁡θ+cos⁡θ1 - \frac{\sin^{3}\theta + \cos^{3}\theta}{\sin\theta + \cos\theta} = 1 - \frac{\left( \sin\theta + \cos\theta \right)\left( \sin^{2}\theta + \cos^{2}\theta - \sin\theta\cos\theta \right)}{\sin\theta + \cos\theta}

=1- (1-sinθcosθ)

= sinθcosθ