Question

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

Answer 1

Verify L.H.S and R.H.S.

Here, $L.H.S.=R.H.S.$

Let $L.H.S.=\frac{\tan \theta }{1-cot\theta }+\frac{cot\theta }{1-\tan \theta }$

$$\begin{gathered} =\frac{\tan \theta }{1-{\displaystyle \frac{1}{\tan \theta }}}+\frac{{\displaystyle \frac{1}{\tan \theta }}}{1-tan\theta } \\ =\frac{{\tan }^2\theta }{\tan \theta -1}+\frac{1}{\tan \theta (1-\tan \theta )} \end{gathered}$$

Step-2

Now $L.C.M.$method

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

We obey $a^3-b^3=(a-b)(a^2+ab-b^2)$

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

$=\tan \theta +1+cot\theta =R.H.S.$

Hence, $L.H.S.=R.H.S.$