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**Hint**: csc stands for cosec. $ \cos ec\left( x \right) $ is a trigonometric function and is reciprocal of $ \sin \left( x \right) $ . $ \cos ec\left( x \right) $ can be defined as the ratio between hypotenuse and perpendicular to theta ( $ \theta $ ). $ \cot \left( x \right) $ is also a trigonometric function and is reciprocal to $ \tan \left( x \right) $ . $ \cot \left( x \right) $ can be defined as the ratio between base and perpendicular to theta ( $ \theta $ ). Both of these trigonometric functions are odd functions.

**:**

__Complete step-by-step answer__As we already know that both $ \cos ec\left( x \right) $ and $ \cot \left( x \right) $ are odd functions, which means that $ f\left( { - x} \right) = - f\left( x \right) $ . So, we can say that,

$ \Rightarrow \csc \left( { - x} \right) = - \csc \left( x \right) $ ---(1)

$ \Rightarrow \cot \left( { - x} \right) = - \cot \left( x \right) $ ---(2)

We are given that $ \dfrac{{\csc \left( { - x} \right)}}{{\cot \left( { - x} \right)}} $ ,

Using equation (1) and (2) in given question we get,

$ \Rightarrow \dfrac{{\csc \left( x \right)}}{{\cot \left( x \right)}} $ ---(3)

We know that $ \cos ec\left( x \right) $ is reciprocal of $ \sin \left( x \right) $ which means

$ \Rightarrow \csc \left( x \right) = \dfrac{1}{{\sin \left( x \right)}} $ ---(4)

And $ \cot \left( x \right) $ is the reciprocal of $ \tan \left( x \right) $ which means

$ \Rightarrow \cot \left( x \right) = \dfrac{{\cos \left( x \right)}}{{\sin \left( x \right)}} $ ---(5)

Now, substituting the values of equation (4) and (5) in equation (3), we get,

$

\Rightarrow \dfrac{{\dfrac{1}{{\sin \left( x \right)}}}}{{\dfrac{{\cos \left( x \right)}}{{\sin \left( x \right)}}}} \\

\Rightarrow \dfrac{1}{{\sin \left( x \right)}} \times \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}} \;

$

$ \sin \left( x \right) $ cuts and cancels off and we get,

$ \Rightarrow \dfrac{1}{{\cos \left( x \right)}} $ ---(6)

We know that, $ \sec \left( x \right) $ is the reciprocal of $ \cos \left( x \right) $ which means that, $ \sec \left( x \right) = \dfrac{1}{{\cos \left( x \right)}} $ . Substituting the same in equation (6) and we get,

$ \Rightarrow \sec \left( x \right) $

Thus, $ \dfrac{{\csc \left( { - x} \right)}}{{\cot \left( { - x} \right)}} = \sec \left( x \right) $ .

**So, the correct answer is “Sec x”.**

**Note**: Before solving any trigonometric question, students should keep all the necessary trigonometric identities in their mind. This will help them to solve any question easily. Students should also remember that $ \cos ec\left( x \right) $ is not the inverse of $ \sin \left( x \right) $ and cannot be written as $ {\sin ^{ - 1}}x $ . The same applies to $ \sec \left( x \right) $ as well as $ \cot \left( x \right) $ . Students should keep in mind that $ \sec \left( x \right) $ is the reciprocal of $ \cos \left( x \right) $ , $ \cos ec\left( x \right) $ is the reciprocal of $ \sin \left( x \right) $ and $ \cot \left( x \right) $ is the reciprocal of $ \tan \left( x \right) $ . In addition to $ \cos ec\left( x \right) $ and $ \cot \le

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