Prove that: \dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec  \theta\ {cosec}\  \theta

L.H.S

$\Rightarrow \frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

$\Rightarrow \frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}$

$\Rightarrow \frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta -\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta -\sin \theta }{\cos \theta }}$

$\Rightarrow \frac{{\sin }^2\theta }{\cos \theta \left(\sin \theta -\cos \theta \right)}+\frac{{\cos }^2\theta }{\sin \theta \left(\cos \theta -\sin \theta \right)}$

$\Rightarrow \frac{{\sin }^2\theta }{\cos \theta \left(\sin \theta -\cos \theta \right)}-\frac{{\cos }^2\theta }{\sin \theta \left(\sin \theta -\cos \theta \right)}$

$\Rightarrow \frac{{\sin }^3\theta -{\cos }^3\theta }{\cos \theta \sin \theta \left(\sin \theta -\cos \theta \right)}$

$\Rightarrow \frac{\left(\sin \theta -\cos \theta \right)\left({\sin }^2+{\cos }^2\theta +\sin \theta \cos \theta \right)}{\cos \theta \sin \theta \left(\sin \theta -\cos \theta \right)}.....\left[\because x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\right]$

$\Rightarrow \frac{\left(1+\sin \theta \cos \theta \right)}{\cos \theta \sin \theta }$

$\Rightarrow 1+\sec \theta \csc \theta$

Hence, proved.