Trigonometric identity: $\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$ I have to prove the following result :
$$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$
I tried converting $\tan\theta$ & $\cot\theta$ into $\cos\theta$ and $\sin\theta$.
That led to a huge expression which I wasn't able to simplify.
Please help!!!
 A: You are on the right track.
writing $\tan\theta$ as$ \dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $ \dfrac {\cos\theta}{\sin\theta} $, we get
$ \dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$
$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)
$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$
$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)
$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)
which is 
$1+\sec\theta\cdot\csc\theta$
QED.
A: $$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$$
$$=\frac{\tan^2\theta}{\tan\theta-1}+\frac{\cot\theta}{1-\tan\theta}(\text{ multiplying the first term by }\tan\theta )$$
$$=-\frac{\tan^2\theta}{1-\tan\theta}+\frac{\cot\theta}{1-\tan\theta}$$
$$=\frac{\cot\theta-\tan^2\theta}{1-\tan\theta}$$
$$=\frac{1-\tan^3\theta}{\tan\theta(1-\tan\theta)}$$
$$=\frac{1+\tan\theta+\tan^2\theta}{\tan\theta}(\text{ assuming }1-\tan\theta\ne0)$$
$$=1+\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$$
$$=1+ \frac1{\sin\theta\cos\theta}$$
A: Just check once again. To begin with, it is correct to put $tan(\theta) = s/c$
( Using obvious abbreviations)
$$\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}
=\frac{s/c}{1-c/s}+\frac{c/s}{1-s/c}$$
$$=\frac{s^2/c -c^2/s}{s-c} =\frac{s^3 - c^3}{s.c.(s-c)}=\frac{s^2+c^2+s.c}{s.c}$$
$$=\frac{1+s.c}{s.c}=1+ 1/{(s.c)} = 1 + {\sec\theta}.{\csc\theta}$$
