Prove the following statement:
 ${{\left( \sin \alpha +\operatorname{cosec}\alpha \right)}^{2}}+{{\left( \cos \alpha +\sec \alpha \right)}^{2}}={{\tan }^{2}}\alpha +{{\cot }^{2}}\alpha +7$.

Hint: In order to prove the relation of the given statement, we need to know certain trigonometric identities like, ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1,{{\tan }^{2}}\alpha +1={{\sec }^{2}}\alpha $ and ${{\cot }^{2}}\alpha +1={{\operatorname{cosec}}^{2}}\alpha $. Also, we should know few algebraic identities too like, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. By using these properties, we can prove the given relation.
Complete step-by-step answer:
In this question, we have been asked to prove that, ${{\left( \sin \alpha +\operatorname{cosec}\alpha \right)}^{2}}+{{\left( \cos \alpha +\sec \alpha \right)}^{2}}={{\tan }^{2}}\alpha +{{\cot }^{2}}\alpha +7$. In order to do that, we will first consider the left hand side or the LHS of the equation. So, we can write it as follows,
$LHS={{\left( \sin \alpha +\operatorname{cosec}\alpha \right)}^{2}}+{{\left( \cos \alpha +\sec \alpha \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we will apply that in the equation and write ${{\left( \sin \alpha +\operatorname{cosec}\alpha \right)}^{2}}={{\sin }^{2}}\alpha +{{\operatorname{cosec}}^{2}}\alpha +2\sin \alpha \operatorname{cosec}\alpha $ and ${{\left( \cos \alpha +sec\alpha \right)}^{2}}={{\cos }^{2}}\alpha +se{{c}^{2}}\alpha +2\cos \alpha sec\alpha $. Now, we will write it in the equation of the LHS,
$LHS={{\sin }^{2}}\alpha +{{\operatorname{cosec}}^{2}}\alpha +2\sin \alpha \operatorname{cosec}\alpha +{{\cos }^{2}}\alpha +se{{c}^{2}}\alpha +2\cos \alpha sec\alpha $
Now, we know that $\sin \alpha =\dfrac{1}{\operatorname{cosec}\alpha }$ and $\cos \alpha =\dfrac{1}{sec\alpha }$. So, if we cross multiply both the equations, we will get, $\sin \alpha \operatorname{cosec}\alpha =1$ and $\cos \alpha \sec \alpha =1$. So, substituting this in the equation of LHS, we get,
$\begin{align}
  & LHS={{\sin }^{2}}\alpha +{{\operatorname{cosec}}^{2}}\alpha +2\left( 1 \right)+{{\cos }^{2}}\alpha +se{{c}^{2}}\alpha +2\left( 1 \right) \\
 & \Rightarrow LHS={{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\operatorname{cosec}}^{2}}\alpha +se{{c}^{2}}\alpha +4 \\
\end{align}$
Now, we also know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$, so applying that in the above equation, we get LHS as,
$\begin{align}
  & LHS=1+{{\operatorname{cosec}}^{2}}\alpha +{{\sec }^{2}}\alpha +4 \\
 & \Rightarrow LHS={{\sec }^{2}}\alpha +{{\operatorname{cosec}}^{2}}\alpha +5 \\
\end{align}$
Now, we know that ${{\tan }^{2}}\alpha +1={{\sec }^{2}}\alpha $ and ${{\cot }^{2}}\alpha +1={{\operatorname{cosec}}^{2}}\alpha $. So, we can write ${{\sec }^{2}}\alpha =1+{{\tan }^{2}}\alpha $ and ${{\operatorname{cosec}}^{2}}\alpha =1+{{\cot }^{2}}\alpha $. So, by substituting it in the equation, we get LHS as,
$LHS=1+{{\tan }^{2}}\alpha +1+{{\cot }^{2}}\alpha +5$
We can also write the LHS as,
$LHS={{\tan }^{2}}\alpha +co{{t}^{2}}\alpha +7$
Which is the right hand side or the RHS of the statement given in the question, so LHS = RHS.
Hence, we have proved that, ${{\left( \sin \alpha +\operatorname{cosec}\alpha \right)}^{2}}+{{\left( \cos \alpha +\sec \alpha \right)}^{2}}={{\tan }^{2}}\alpha +{{\cot }^{2}}\alpha +7$.

Note: While solving this question, one can think of converting $\sec \alpha $ and $\operatorname{cosec}\alpha $ in $\cos \alpha $ and $\sin \alpha $ respectively and simplifying. It is not wrong to do so, but it will make the question lengthy and complicated. Also, we need to remember that $\sin \alpha =\dfrac{1}{\operatorname{cosec}\alpha }$, which can be written as $\sin \alpha \operatorname{cosec}\alpha =1$ and similarly $\cos \alpha =\dfrac{1}{sec\alpha }$ can be written as $\cos \alpha \sec \alpha =1$.