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**Hint:**We must know about the relations among the different values given by sine and cosine functions on having angles that differ by multiples of $90{}^\circ $. These relations are mentioned below

$\begin{align}

& \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \\

& \sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta \\

& \sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \\

& \sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta \\

& \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\

& \cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta \\

& \cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \\

& \cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta \\

\end{align}$

Now, using these expressions and relations of sine and cosine functions, we will simplify the given equation and find the value of $x$.

**Complete step by step answer:**

We will first simplify the L.H.S. of the given equation by using the relations mentioned above, in the following manner,

L.H.S. $=\cos \left( 270{}^\circ +\theta \right)+x\cdot \cos \theta \cdot \cot \left( 180{}^\circ +\theta \right)$

Now, we know that $\cot \left( {{180}^{\circ }}+\theta \right)=\dfrac{\cos \left( {{180}^{\circ }}+\theta \right)}{\sin \left( {{180}^{\circ }}+\theta \right)}=\dfrac{-\cos \theta }{-\sin \theta }=\dfrac{\cos \theta }{\sin \theta }$.

The value for $\cos \left( {{270}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+{{90}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)$ . We will use the relations mentioned above in the following manner,

$\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)=-\cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta $. Therefore, we have $\cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $.

Substituting these values in the L.H.S., we get

L.H.S. $=\sin \theta +x\cdot \cos \theta \cdot \dfrac{\cos \theta }{\sin \theta }$

Simplifying this equation, we get

L.H.S. $=\sin \theta +x\cdot \dfrac{{{\cos }^{2}}\theta }{\sin \theta }=\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }$

Now, we will look at the R.H.S.,

R.H.S. $=\sin \left( {{270}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)$

Again, we will use the relations mentioned above to simplify the R.H.S. as follows,

$\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)=-\sin \left( {{90}^{\circ }}-\theta \right)=-\cos \theta $

Now, we will equate the L.H.S. and R.H.S. and get the following equation,

$\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }=-\cos \theta $

Simplifying the above equation, we get

${{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta =-\cos \theta \sin \theta $

$\therefore x\cdot {{\cos }^{2}}\theta =-{{\sin }^{2}}\theta -\cos \theta \sin \theta $

Dividing by ${{\cos }^{2}}\theta $ on both sides of the above equation, we get

$\begin{align}

& x=-\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-\dfrac{\cos \theta \sin \theta }{{{\cos }^{2}}\theta } \\

& =-{{\tan }^{2}}\theta -\tan \theta

\end{align}$

**Hence, the value is $x={{\tan }^{2}}\theta -\tan \theta $.**

**Note:**It is easy to make an error in such types of questions, if we are not aware of the relations that are mentioned in the hint as they are very crucial to solve the problem and get to the solution. In this question, we have used multiple relations and trigonometric identities to simplify the given equation. While solving questions related to trigonometry, simplification of the given equations is the key aspect.

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