Prove the following expression
\[{{\left( 1+\tan \alpha \tan \beta \right)}^{2}}+{{\left( \tan \alpha -\tan \beta \right)}^{2}}={{\sec }^{2}}\alpha {{\sec }^{2}}\beta \]

Hint: In this question, we have to prove a relation for which we will require the knowledge of a few trigonometric identities like \[{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \]. Also, we will require the knowledge of a few algebraic identities like \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].

Complete step-by-step answer:

In this question, we have to prove that

\[{{\left( 1+\tan \alpha \tan \beta \right)}^{2}}+{{\left( \tan \alpha -\tan \beta \right)}^{2}}={{\sec }^{2}}\alpha {{\sec }^{2}}\beta \]

To prove this relation we will first consider the terms on the left-hand side. So, we can write it as,

\[LHS={{\left( 1+\tan \alpha \tan \beta \right)}^{2}}+{{\left( \tan \alpha -\tan \beta \right)}^{2}}\]

Now, we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. So, we can write \[{{\left( 1+\tan \alpha \tan \beta \right)}^{2}}=1+{{\tan }^{2}}\alpha {{\tan }^{2}}\beta +2\tan \alpha \tan \beta \] and \[{{\left( \tan \alpha -\tan \beta \right)}^{2}}={{\tan }^{2}}\alpha +{{\tan }^{2}}\beta -2\tan \alpha \tan \beta \]. Therefore, we will get LHS as,

\[LHS=1+{{\tan }^{2}}\alpha {{\tan }^{2}}\beta +2\tan \alpha \tan \beta +{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta -2\tan \alpha \tan \beta \]

Now, we know that the equal terms with opposite signs get canceled out. So, we will get
\[LHS=1+{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta +{{\tan }^{2}}\alpha {{\tan }^{2}}\beta \]

Now we will take \[{{\tan }^{2}}\beta \] common from the term \[{{\tan }^{2}}\beta \] and \[{{\tan }^{2}}\alpha {{\tan }^{2}}\beta \]. So, we will get LHS as

\[LHS=\left( 1+{{\tan }^{2}}\alpha \right)+{{\tan }^{2}}\beta \left( 1+{{\tan }^{2}}\alpha \right)\]

Now, we can also take \[\left( 1+{{\tan }^{2}}\alpha \right)\] common from the LHS, so we will get

\[LHS=\left( 1+{{\tan }^{2}}\alpha \right)\left( 1+{{\tan }^{2}}\beta \right)\]

Now we know that \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]. So, we can write \[1+{{\tan }^{2}}\alpha ={{\sec }^{2}}\alpha \] and \[1+{{\tan }^{2}}\beta ={{\sec }^{2}}\beta \].

Therefore, we will get

\[LHS={{\sec }^{2}}\alpha {{\sec }^{2}}\beta \]

LHS = RHS

Hence proved


Note: We could have also done this question by converting \[\tan \theta \] in terms of \[\sin \theta \] and \[\cos \theta \], that is by putting \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] , and then we could have simplified it. But it would become a complicated and lengthy solution. And lengthier the solution, more the chances of calculation mistake.