Prove the following trigonometric identities.

We need to prove $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$

Now, using $\cot \theta=\frac{1}{\tan \theta}$ in the L.H.S, we get

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\frac{\tan \theta}{\left(1-\frac{1}{\tan \theta}\right)}+\frac{\left(\frac{1}{\tan \theta}\right)}{1-\tan \theta}$

$=\frac{\tan \theta}{\left(\frac{\tan \theta-1}{\tan \theta}\right)}+\frac{1}{\tan \theta(1-\tan \theta)}$

$=\left(\frac{\tan \theta}{\tan \theta-1}\right)(\tan \theta)+\frac{1}{\tan \theta(1-\tan \theta)}$

$=\frac{\tan ^{2} \theta}{\tan \theta-1}-\frac{1}{\tan \theta(\tan \theta-1)}$

$=\frac{\tan ^{3} \theta-1}{\tan \theta(\tan \theta-1)}$

Further using the identity $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, we get

$\frac{\tan ^{3} \theta-1}{\tan \theta(\tan \theta-1)}=\frac{(\tan \theta-1)\left(\tan ^{2} \theta+\tan \theta+1\right)}{\tan \theta(\tan \theta-1)}$

$=\frac{\tan ^{2} \theta+\tan \theta+1}{\tan \theta}$

$=\frac{\tan ^{2} \theta}{\tan \theta}+\frac{\tan \theta}{\tan \theta}+\frac{1}{\tan \theta}$

$=\tan \theta+1+\cot \theta$

Hence $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$