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**Hint**: This is a very simple question from the chapter of trigonometry. As we can see, no option is in $ \sin x $ or $ \sec x $ . The options have one or combination of trigonometric ratios of $ \tan x $ , $ \cot x $ and $ \cos \text{ec}x $. This means we have to use properties of trigonometric ratios and manipulate the given trigonometric expression $ \sin x\sec x $ so that it matches one of the options.

**:**

__Complete step-by-step answer__To solve this question, we will define the various trigonometric ratios.

The sine or sin of an angle is defined as the ratio of the side opposite to the angle x and the hypotenuse of the right angled triangle.

Thus, $ \sin x=\dfrac{opp}{hyp} $

cosine or cos of an angle is defined as the ratio of the side adjacent to the angle x and the hypotenuse of the right angled triangle.

Thus, $ \cos x=\dfrac{adj}{hyp} $

tangent or tan of an angle is defined as the ratio of the side opposite to the angle x and the side adjacent to the angle x of the right angled triangle.

Thus, $ \tan x=\dfrac{opp}{adj} $

$ \Rightarrow \tan x=\dfrac{\sin x}{\cos x} $

cotangent or cot of an angle is defined as the ratio of the side adjacent to the angle x and the side opposite of the angle x of the right angled triangle. Thus, it is reciprocal of $ \tan x $ .

Therefore, $ \cot x=\dfrac{adj}{opp} $

$ \Rightarrow \cot x=\dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x} $

secant or sec of an angle is defined as the ratio of the hypotenuse and the side adjacent to angle x of the right angled triangle. Thus, it is reciprocal of $ \cos x $ .

Therefore, $ \sec x=\dfrac{hyp}{adj} $

$ \Rightarrow \sec x=\dfrac{1}{\cos x} $

cosecant or cosec of an angle is defined as the ratio of the hypotenuse and the side opposite to angle x of the right angled triangle. Thus, it is reciprocal of $ \sin x $ .

Therefore, $ \cos \text{ec}x=\dfrac{hyp}{opp} $

$ \Rightarrow \cos \text{ec}x=\dfrac{1}{\sin x} $

The expression given to us is $ \sin x\sec x $ . In this expression, we will replace $ \sec x $ with $ \dfrac{1}{\cos x} $ .

$ \Rightarrow \dfrac{\sin x}{\cos x} $

But we know that $ \dfrac{\sin x}{\cos x} $ is $ \tan x $ .

$ \Rightarrow \dfrac{\sin x}{\cos x}=\tan x $

**So, the correct answer is “Option A”.**

**Note**: Students are advised to be well versed in the concepts of trigonometry. This is a fairly simple problem if the students know the basic properties of trigonometric ratios. The reciprocal functions must be known to solve this problem.

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