Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$ I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$ 
a)$1-\tan\theta-\cot\theta$
b)$1+\tan\theta-\cot\theta$
c)$1-\tan\theta+\cot\theta$
d)$1+\tan\theta+\cot\theta$
I've tried it several ways like taking LCM,change whole into $\sin\theta$ and $\cos\theta$.but  I've stuck.

I'm not sure where you got stuck after writing the expression in terms of $ \sin \theta $ and $ \cos \theta $ but here's how I would do it: $$ \begin{align*}\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} &= \frac{\sin \theta \tan \theta}{\sin \theta - \cos \theta} + \frac{\cos\theta \cot \theta}{\cos \theta - \sin \theta} \\ &= \frac{\sin\theta\tan\theta - \cos\theta \cot \theta}{\sin \theta - \cos \theta} \\&= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin\theta - \cos\theta)} \\&= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2\theta }{\sin \theta \cos \theta}\\&= 1 + \tan \theta + \cot \theta\\&=\boxed{\text{D}} \end{align*} $$
