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.
$$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=\frac{\tan\theta}{1-\frac{1}{\tan\theta}}+\frac{\cot\theta}{1-\tan\theta}=$$
$$=\frac{-\tan^2\theta}{1-{\tan\theta}}+\frac{\cot\theta}{1-\tan\theta}=\frac{\frac{1}{\tan\theta}-\tan^2\theta}{1-{\tan\theta}}=$$
$$=\frac{1-\tan^3\theta}{\tan\theta(1-{\tan\theta})}=\frac{(1-\tan\theta)(1+\tan\theta+\tan^2\theta)}{\tan\theta(1-{\tan\theta})}=$$
$$=\frac{1+\tan\theta+\tan^2\theta}{\tan\theta}=1+\cot\theta+\tan\theta$$
Just to add the "cheating" method of solving these (say on an exam when you're pressed for time):
Use the symmetry to show that only options (a) and (d) are viable. Then plug in some value (say $x=\pi/8$), to find which one is of them is "right"
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: $$ \begin{align*}\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} &= \frac{\sin \theta \tan \theta}{\sin \theta - \cos \theta} + \frac{\cos\theta \cot \theta}{\cos \theta - \sin \theta} \\ &= \frac{\sin\theta\tan\theta - \cos\theta \cot \theta}{\sin \theta - \cos \theta} \\&= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin\theta - \cos\theta)} \\&= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2\theta }{\sin \theta \cos \theta}\\&= 1 + \tan \theta + \cot \theta\\&=\boxed{\text{D}} \end{align*} $$
$$ \begin{align} \frac{\tan(\theta)}{1-\cot(\theta)}+\frac{\cot(\theta)}{1-\tan(\theta)} &=\frac{\cot(\theta)-\tan^2{\theta}}{1-\tan(\theta)}\\ &=\cot(\theta)\frac{1-\tan^3(\theta)}{1-\tan(\theta)}\\ &=\cot(\theta)+1+\tan(\theta) \end{align} $$
