The expression \[\dfrac{{\cos ({{90}^ \circ } + \theta )\sec ( - \theta )\tan ({{180}^ \circ } - \theta )}}{{\sin ({{360}^ \circ } + \theta )\sec ({{180}^ \circ } + \theta )\cot ({{90}^ \circ } - \theta )}}\] is equal to ?
A. 2
B. 1
C. -1
D. 0

Hint:Here, the question is related to the trigonometry topic. By using the trigonometric function for allied angles, we write the value for the respective trigonometric function for the allied angles. On further simplification we obtain the required answer for the given question.

Complete step by step answer:
In trigonometry we have six trigonometric ratios namely, sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The angles will be present in any of the fourth quadrants. The sign of the trigonometric ratio will be based on the ASTC rule.

If an angle is multiple of \[{90^ \circ }\], then \[\sin \Leftrightarrow \cos \], \[\tan \Leftrightarrow \cot \] and \[\sec \Leftrightarrow \csc \]. If an angle is multiple of \[{180^ \circ }\], then \[\sin \Leftrightarrow \sin \], \[\tan \Leftrightarrow \tan \] and \[\sec \Leftrightarrow \sec \]. When the angle is multiple of \[{90^ \circ }\], then it will be converted into the co-ratios of trigonometry. This is called a trigonometric function for allied angles.

The other trigonometric ratios for allied angles is given below:
When the angle is a negative
\[\sin ( - \theta ) = - \sin \theta \\
\Rightarrow \cos ( - \theta ) = \cos \theta \\
\Rightarrow \tan ( - \theta ) = - \tan \theta \\
\Rightarrow \csc ( - \theta ) = - \csc \theta \\
\Rightarrow \sec ( - \theta ) = \sec \theta \\
\Rightarrow \cot ( - \theta ) = - \cot \theta \\ \]
When the angle is the sum of \[{90^ \circ }\]
\[\sin ({90^ \circ } + \theta ) = \cos \theta \\
\Rightarrow \cos ({90^ \circ } + \theta ) = - \sin \theta \\
\Rightarrow \tan ({90^ \circ } + \theta ) = - \cot \theta \\
\Rightarrow \csc ({90^ \circ } + \theta ) = \sec \theta \\
\Rightarrow \sec ({90^ \circ } + \theta ) = - \csc \theta \\
\Rightarrow \cot ({90^ \circ } + \theta ) = - \tan \theta \\ \]
When the angle is the difference of \[{90^ \circ }\]
\[\sin ({90^ \circ } - \theta ) = \cos \theta \\
\Rightarrow \cos ({90^ \circ } - \theta ) = \sin \theta \\
\Rightarrow \tan ({90^ \circ } - \theta ) = \cot \theta \\
\Rightarrow \csc ({90^ \circ } - \theta ) = \sec \theta \\
\Rightarrow \sec ({90^ \circ } - \theta ) = \csc \theta \\
\Rightarrow \cot ({90^ \circ } - \theta ) = \tan \theta \\ \]
When the angle is the sum of \[{180^ \circ }\]
\[\sin ({180^ \circ } + \theta ) = - \sin \theta \\
\Rightarrow \cos ({180^ \circ } + \theta ) = - \cos \theta \\
\Rightarrow \tan ({180^ \circ } + \theta ) = \tan \theta \\
\Rightarrow \csc ({180^ \circ } + \theta ) = - \csc \theta \\
\Rightarrow \sec ({180^ \circ } + \theta ) = - \sec \theta \\
\Rightarrow \cot ({180^ \circ } + \theta ) = \cot \theta \\ \]
When the angle is the difference of \[{180^ \circ }\]
\[\sin ({180^ \circ } - \theta ) = \sin \theta \\
\Rightarrow \cos ({180^ \circ } - \theta ) = - \cos \theta \\
\Rightarrow \tan ({180^ \circ } - \theta ) = - \tan \theta \\
\Rightarrow \csc ({180^ \circ } - \theta ) = \csc \theta \\
\Rightarrow \sec ({180^ \circ } - \theta ) = - \sec \theta \\
\Rightarrow \cot ({180^ \circ } - \theta ) = - \cot \theta \\ \]
When the angle is the sum of \[{360^ \circ }\]
\[\sin ({360^ \circ } + \theta ) = \sin \theta \\
\Rightarrow \cos ({360^ \circ } + \theta ) = \cos \theta \\
\Rightarrow \tan ({360^ \circ } + \theta ) = \tan \theta \\
\Rightarrow \csc ({360^ \circ } + \theta ) = \csc \theta \\
\Rightarrow \sec ({360^ \circ } + \theta ) = \sec \theta \\
\Rightarrow \cot ({360^ \circ } + \theta ) = \cot \theta \\ \]
Now we consider the given question and simplify it
\[\dfrac{{\cos ({{90}^ \circ } + \theta )\sec ( - \theta )\tan ({{180}^ \circ } - \theta )}}{{\sin ({{360}^ \circ } + \theta )\sec ({{180}^ \circ } + \theta )\cot ({{90}^ \circ } - \theta )}}\]
By the trigonometric function for allied angles the above inequality is written as
\[\dfrac{{\sin \theta \times \sec \theta \times - \tan \theta }}{{\sin \theta \times \sec \theta \times \tan \theta }}\]
This can be written as
\[ \Rightarrow \dfrac{{ - (\sin \theta \times \sec \theta \times \tan \theta )}}{{\sin \theta \times \sec \theta \times \tan \theta }}\]
On cancelling the terms in the numerator and denominator we have
\[ \Rightarrow - 1\]
Therefore \[\dfrac{{\cos ({{90}^ \circ } + \theta )\sec ( - \theta )\tan ({{180}^ \circ } - \theta )}}{{\sin ({{360}^ \circ } + \theta )\sec ({{180}^ \circ } + \theta )\cot ({{90}^ \circ } - \theta )}} = - 1\]

Hence the option C is the correct answer.

Note:The ASTC rule is defined as “All Sine Cosine Tangent”. In the first quadrant all, in a second quadrant sine, in a third quadrant cosine and in a fourth quadrant tangent respectively trigonometric ratios are positive. The Allied angles will be based on the ASTC rule.