Question

If $\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }-\frac{\cos \theta }{\sqrt{1+{\cot }^2\theta }}-2\tan \theta \cdot \cot \theta =-1$ where $\theta \in [0,2\pi ]$ then

Answer 1

$\begin{array}{c}\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }+\frac{\cos \theta }{\sqrt{1+{\cos }^2\theta }}-2\tan \theta \cot \theta =-1\\ \frac{\left(\sin \theta -\cos \theta \right)\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta \right)}{\left(\sin \theta -\cos \theta \right)}+\frac{\cos \theta }{\cos ec\theta }\\ -2=-1\\ -1+\sin \theta .\cos \theta +\sin \theta \cos \theta =-1\\ 2\sin \theta .\cos \theta =0\\ \sin 2\theta =0\\ 2\theta =n\pi \\ \theta =\frac{n\pi }{2}\\ \theta =\frac{\pi }{2}or\pi or\frac{3\pi }{2}\end{array}$