Find the value of the trigonometric expression $\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1$.

Hint: We will use the trigonometric angular formula which are given by $\cos \left( 72 \right)=\cos \left( 90-18 \right)$, $\cos \left( 90-18 \right)=\sin \left( 18 \right)$, $\text{cosec}\left( 90-15 \right)=\sec \left( 15 \right)$ and $\cot \left( 18 \right)=\cot \left( 90-72 \right)$ to solve the given trigonometric question. By this we will be able to reduce any angle in terms of 90 degrees and after that we will cancel the common terms.

Complete step-by-step answer:
We will consider the equation $\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1$...(i).
First we will consider the left hand side of the expression we will have $\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}$.
Now we will consider the expression$\cos \left( 72 \right)=\cos \left( 90-18 \right)$. As we know that the value of $\cos \left( 90-18 \right)=\sin \left( 18 \right)$ therefore, we will get a new expression which is given by,
$\begin{align}
  & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\sin \left( 18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\
 & \Rightarrow \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\
\end{align}$
Also if we put the value of the trigonometric term $\text{cosec}\left( 90-15 \right)=\sec \left( 15 \right)$ in the term of angle as 15, therefore we now have the trigonometric expression as,
$\begin{align}
  & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\
 & \Rightarrow \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\text{cosec}\left( 90-15 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\
 & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\sec \left( 15 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)} \\
 & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 18 \right)} \\
\end{align}$
Also, as we know that the value of $\cot \left( 18 \right)=\cot \left( 90-72 \right)$ thus we will get
$\begin{align}
  & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 18 \right)} \\
 & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\cot \left( 90-72 \right)} \\
 & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=\dfrac{\tan \left( 72 \right)}{\tan \left( 72 \right)} \\
 & \dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 90-18 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1 \\
\end{align}$
Hence, we have proved the desired trigonometric equation which is given by $\dfrac{\sin \left( 18 \right)+\tan \left( 72 \right)-\cos \left( 72 \right)}{\text{cosec}\left( 75 \right)+\cot \left( 18 \right)-\sec \left( 15 \right)}=1$.

Note: We can also substitute the expression $\sin \left( 18 \right)=\sin \left( 90-72 \right)$ in the expression and proceed in the same way. This will also result in the right answer. We need to take care that as we are taking the angles in terms of 90 degrees so the trigonometric terms will be the same. For example of $\cot \left( 18 \right)=\cot \left( 90-72 \right)$instead of $\cot \left( 18 \right)=\tan \left( 90-72 \right)$. We will consider the left hand side first of the given trigonometric expression so that we will lead the left hand side to the right side of the expression. By considering these points in notice we will get the desired result.