Question

$\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }-\frac{\cos \theta }{\sqrt{1+{\cot }^2\theta }}-2\tan \theta \cot \theta =-1$ , is true for

• A. $\theta ϵ(0,\frac{\pi }{2})$
• B. $\theta ϵ\left(\frac{\pi }{2},\pi \right)$
• C. $\theta ϵ(\pi ,\frac{3\pi }{2})$
• D. $\theta ϵ\left(\frac{3\pi }{2},2\pi \right)$

Answer 1

Answer: (A), (B)

$\theta ϵ(0,\frac{\pi }{2})$
$\theta ϵ\left(\frac{\pi }{2},\pi \right)$

We have, $\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }={\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta$
Using this in the given equation we get,
$\Rightarrow 1+\sin \theta \cos \theta -|\sin \theta |\cos \theta -2=-1$
$\Rightarrow |\sin \theta |\cos \theta =\sin \theta \cos \theta$
So both option A and B are appropriate