 QUESTION

# Prove that $\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta}}{{1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta$

Hint: Make use of the definition of trigonometric ratios that is of $\tan \theta$ and
$\cot \theta$ and solve it.

We have been given with
LHS=$\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}$
Now let us express $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
So, we get $\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}$
Now, let us take LCM for the terms in the denominator and solve it
So, we get $\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}$

On solving further,$\dfrac{{{{\sin }^2}\theta }}{{\cos \theta (\sin \theta - \cos \theta )}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta (\cos \theta - \sin \theta )}}$
$\Rightarrow \dfrac{{{{\sin }^2}\theta }}{{\cos \theta (\sin \theta - \cos \theta )}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta (\sin \theta - \cos \theta )}}$
Now, let us take out $\dfrac{1}{{\sin \theta - \cos \theta }}$ common from the denominator
So, we get
$\dfrac{1}{{\sin \theta - \cos \theta }}\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right] \\ \\$

On taking LCM and solving further, we get
$\dfrac{1}{{\sin \theta - \cos \theta }}\left[ {\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\sin \theta \cos \theta }}} \right]$
Now ${\sin ^3}\theta - {\cos ^3}\theta$ is of the form ${a^3} - {b^3}$
We know ${a^3} - {b^3} = (a - b)({a^2} + {b^2} +ab)$
So, now we can write in the equation ${\sin ^3}\theta - {\cos ^3}\theta$ as
$\dfrac{1}{{\sin \theta - \cos \theta }}\left[ {\dfrac{{\left( {\sin \theta - \cos \theta } \right)({{\sin }^2} + {{\cos }^2}\theta + \sin \theta \cos \theta )}}{{\sin \theta \cos \theta }}} \right]$
${\sin ^2}\theta + {\cos ^2}\theta = 1$
So, on making use of this in the equation ,we get
$\dfrac{{\left( {1 + \sin \theta \cos \theta } \right)}}{{\sin \theta \cos \theta }} \\ = \dfrac{1}{{\sin \theta \cos \theta }} + 1 \\ \dfrac{1}{{\sin \theta }} = \cos ec\theta ,\dfrac{1}{{\cos \theta }} = \sec \theta \\$
So, we get $1 + \sec \theta \cos ec\theta$ =RHS,
Hence the result is proved.

Note: We have to modify the equations here in accordance to the RHS which we have
to prove and also make use of the suitable trigonometric identities to wherever needed
so that the equations can be simplified.