Solve the expression $\dfrac{\cos (90-\theta )\sec (90-\theta )\tan \theta }{\csc (90-\theta )\sin (90-\theta )\cot (90-\theta )}+\dfrac{\tan (90-\theta )}{\cot \theta }=...........$
$A.$ $1$
$B.$ $-1$
$C.$ $2$
$D.$ $-2$

Hint: All the values of $\sin \theta ,\cos \theta ,\tan \theta $ are positive in the first quadrant. So, we will directly use the following formulae to evaluate.

$\begin{align}

& \cos (90-\theta )=\sin \theta ,\sec (90-\theta )=\csc \theta \\

& \csc (90-\theta )=\sec \theta ,\sin (90-\theta )=\cos \theta \\

& \cot (90-\theta )=\tan \theta ,\tan (90-\theta )=\cot \theta \\

\end{align}$

Complete step-by-step answer:
It is given in the question that to evaluate the value of

$\dfrac{\cos (90-\theta )\sec (90-\theta )\tan \theta }{\csc (90-\theta )\sin (90-\theta )\cot (90-\theta )}+\dfrac{\tan (90-\theta )}{\cot \theta }=$…………

We know that all the values of $\sin \theta ,\cos \theta ,\sec \theta ,\csc \theta ,\cot \theta ,\tan \theta $ are positive in the first quadrant.


We also know some basic formulas of trigonometry like,

$\begin{align}
& \cos (90-\theta )=\sin \theta ,\sec (90-\theta )=\csc \theta \\
& \csc (90-\theta )=\sec \theta ,\sin (90-\theta )=\cos \theta \\
& \cot (90-\theta )=\tan \theta ,\tan (90-\theta )=\cot \theta \\
\end{align}$
We will use all these trigonometric formulas to evaluate the given expression.
So, on placing the values of $\cos (90-\theta )$ as $\sin \theta $, $\sec (90-\theta )$ as $\csc \theta $, $\csc (90-\theta )$ as $\sec \theta $, $\sin (90-\theta )$ as $\cos \theta $, $\cot (90-\theta )$ as $\tan \theta $, $\tan (90-\theta )$ as $\cot \theta $ in expression, we get

$\dfrac{\cos (90-\theta )\sec (90-\theta )\tan \theta }{\csc (90-\theta )\sin (90-\theta )\cot (90-\theta )}+\dfrac{\tan (90-\theta )}{\cot \theta }=\dfrac{\sin \theta .\csc \theta .\tan \theta }{\sec \theta .\cos \theta .\tan \theta }+\dfrac{\cot \theta }{\cot \theta }$

Cancelling $\cot \theta $ and $\tan \theta $ from numerator and denominator, we get

$=\dfrac{\sin \theta .\csc \theta .1}{\sec \theta .\cos \theta .1}+\dfrac{1}{1}$

$=\dfrac{\sin \theta .\csc \theta }{\sec \theta .\cos \theta }+1$

We know that,

$\csc \theta =\dfrac{1}{\sin \theta }$ And $\sec \theta =\dfrac{1}{\cos \theta }$

Placing the values of $\csc \theta $ as $\dfrac{1}{\sin \theta }$ and $\sec \theta $ as $\dfrac{1}{\cos \theta }$ in the expression, we get

$=\dfrac{\sin \theta \times \dfrac{1}{\sin \theta }}{\dfrac{1}{\cos \theta }\times \cos \theta }+1$

Cancelling similar terms from the above expression, we get

$\begin{align}

& =\dfrac{1}{1}+1 \\

& =1+1 \\

& =2 \\

\end{align}$

Thus, we get the value of expression is 2 and option $C$ is the correct answer.

Note: This question can be solved in just a few steps if you know these basic formulas of trigonometry. Without the use of these formulas your solution becomes complex and chances of error will also increase. You can also solve this question by using graphs of all the trigonometric values but, the process is quite complex and evaluating all the angles and graphs without any error is not worth it for this question.