State whether the given is true or false.
$\dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}-2\cos ({{135}^{\circ }}+\alpha )\cos ({{315}^{\circ }}-\alpha )=2\cos 2\alpha $
$A)True$
$B)False$

Hint : To solve the question we need to have the knowledge of trigonometric identities. In this question we need to convert all the identities into different identities to prove the Right Hand Side of the expression. The concept used here will be that all the trigonometric functions is positive for the angles present in first quadrant, $\sin $and $\text{cosec}$ is a positive in second quadrant, $\cot $and $\tan $ is a positive in third quadrant while $\cos $and $\text{sec}$ is a positive for the angle in fourth quadrant.

Complete step-by-step solution:
The question ask us to check whether the expression given $\dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}-2\cos ({{135}^{\circ }}+\alpha )\cos ({{315}^{\circ }}-\alpha )=2\cos 2\alpha $ is correct or not. We will consider the left hand side of the equation which is in fraction form. We will firstly convert$\cot \alpha $ into a function which will be much easier for us to work upon.
The first step will be to convert $\cot \left( {{270}^{\circ }}-\alpha \right)$ into the trigonometric function with a single angle. For that we will convert$\cot \left( {{270}^{\circ }}-\alpha \right)$ into $\tan $. Since the angle$\left( {{270}^{\circ }}-\alpha \right)$lies in the third quadrant and ${{270}^{\circ }}$ is thrice of ${{90}^{\circ }}$. So the function $\cot \left( {{270}^{\circ }}-\alpha \right)$ will be converted into$\tan \alpha $, as $\cot $ is positive in the third quadrant. So on writing this we get:
$\Rightarrow \dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}=\dfrac{\cot \alpha +\tan \alpha }{\cot \alpha -\tan \alpha }$
On dividing the fraction by $\cot \alpha $ we get:
\[\Rightarrow \dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}=\dfrac{\dfrac{\cot \alpha +\tan \alpha }{\cot \alpha }}{\dfrac{\cot \alpha -\tan \alpha }{\cot \alpha }}\]
\[\Rightarrow \dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}=\dfrac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }\]
Since the formula \[\dfrac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }\]is equal to $\cos 2\alpha $ . So the above expression turns to:
\[\Rightarrow \dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}=\dfrac{1+{{\tan }^{2}}\alpha }{1-{{\tan }^{2}}\alpha }=\cos 2\alpha \]……….(i)
Now we will work upon the other part of the expression present in Left Hand Side.
$\Rightarrow 2\cos ({{135}^{\circ }}+\alpha )\cos ({{315}^{\circ }}-\alpha )=2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)$
$\Rightarrow 2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)=-2\sin \left( {{45}^{\circ }}+\alpha \right)\cos \left( {{45}^{\circ }}+\alpha \right)$
$\Rightarrow 2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)=-2\sin \left( {{45}^{\circ }}+\alpha \right)\cos \left( {{45}^{\circ }}+\alpha \right)$
We will use the formula $2\sin x\cos x=\sin 2x$ in the above expression:
$\Rightarrow 2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)=-\sin \left( 2\left( {{45}^{\circ }}+\alpha \right) \right)$
$\Rightarrow 2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)=-\sin \left( {{90}^{\circ }}+2\alpha \right)$
$\Rightarrow 2\cos \left( {{90}^{\circ }}+\left( {{45}^{\circ }}+\alpha \right) \right)\cos \left( {{360}^{\circ }}-\left( {{45}^{\circ }}+\alpha \right) \right)=-\cos 2\alpha $………..(ii)
On substituting the equation (i) and (ii) in the question we will get:
$\Rightarrow \dfrac{\cot \alpha +\cot \left( {{270}^{\circ }}-\alpha \right)}{\cot \alpha -\cot \left( {{270}^{\circ }}-\alpha \right)}-2\cos ({{135}^{\circ }}+\alpha )\cos ({{315}^{\circ }}-\alpha )$
$\Rightarrow \cos 2\alpha -(-\cos 2\alpha )$
$\Rightarrow \cos 2\alpha +\cos 2\alpha $
$\Rightarrow 2\cos 2\alpha $
So the above expression is correct.
$\therefore $ The above expression is $A)True$ .

Note: To solve this question we need to remember the quadrant in which the particular trigonometric function is positive or negative, as it helps in conversion of one trigonometric function to another. We need to remember the other trigonometric formulas too.