Question

Prove that $ {{\left( \tan A-\tan B \right)}^{2}}+{{\left( 1+\tan A\tan B \right)}^{2}}={{\sec }^{2}}A{{\sec }^{2}}B $

Answer 1

Hint: In this question, we have to use the below formulae for the simplification of the given trigonometric equation.
 $ \begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align} $
Below trigonometric identity is also used to solve the above trigonometric equation.
 $ {{\sec }^{2}}A-{{\tan }^{2}}A=1 $
By using the above formulae we should try to solve the L.H.S of the equation and it should be equal to the R.H.S. of the equation.

Complete step by step answer:
Now considering from the question, we were given to prove is $ {{\left( \tan A-\tan B \right)}^{2}}+{{\left( 1+\tan A\tan B \right)}^{2}}={{\sec }^{2}}A{{\sec }^{2}}B $
To prove the given equation we should equate the L.H.S and R.H.S of the given equation. First of all we should solve the left-hand side of the equation.
 $ L.H.S={{\left( \tan A-\tan B \right)}^{2}}+{{\left( 1+\tan A\tan B \right)}^{2}} $
Now as we have been already discussed above we have to use the below-mentioned basic formulae to solve the given trigonometric equation.
 $ \begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align} $
 $ \begin{align}
  & L.H.S={{\left( \tan A-\tan B \right)}^{2}}+\left( 1+\tan A\tan B \right) \\
 & \Rightarrow {{\tan }^{2}}A+{{\tan }^{2}}B-2\tan A\tan B+1+{{\tan }^{2}}A{{\tan }^{2}}B+2\tan A\tan B
\end{align} $
In the above equation $ 2\tan A\tan B $ and $ -2\tan A\tan B $ gets cancelled.
Now, $ L.H.S=L.H.S=1+{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}A{{\tan }^{2}}B $
Now taking the common terms in the equation we get the below equation
 $ L.H.S=1\left( 1+{{\tan }^{2}}B \right)+{{\tan }^{2}}A\left( {{\tan }^{2}}B+1 \right) $
 $ L.H.S=\left( 1+{{\tan }^{2}}B \right)+\left( 1+{{\tan }^{2}}A \right) $
Now, we all know that we have one trigonometric identity
 $ \begin{align}
  & {{\sec }^{2}}A-{{\tan }^{2}}A=1 \\
 & {{\sec }^{2}}A=1+{{\tan }^{2}}A \\
\end{align} $
By using the above identity we can solve the L.H.S of the given equation.
 $ L.H.S={{\sec }^{2}}B{{\sec }^{2}}A. $
By solving the L.H.S using appropriate formulae we get the above equation as the final equation.
Therefore, L.H.S = R.H.S
Hence it is proved.

Note:
We have to observe the equation thoroughly and should recognize which formula is suitable for solving the given trigonometric equation. After knowing the formulae we have to apply the basic formulae correctly without any mistake. We have to use trigonometric identities carefully. The calculation must be done very carefully while splitting the formulae and cancellation of terms.