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# Prove ${{\text{x}}^2} - {y^2} = {a^2} - {b^2}$, if $x = a\sec \theta + b\tan \theta$ and $y = a\tan \theta + b\sec \theta$.

Last updated date: 13th Jun 2024
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Hint: Here we will substitute the value of x and y on the left hand side of the equation and prove it equal to the right hand side using various trigonometric identities:-
$1 + {\tan ^2}\theta = {\sec ^2}\theta \\ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\$
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

The given equation is:-
${{\text{x}}^2} - {y^2} = {a^2} - {b^2}$
Let us consider the left hand side of the given equation:-
$LHS = {{\text{x}}^2} - {y^2}$
Now it is given that:-
$x = a\sec \theta + b\tan \theta$ and $y = a\tan \theta + b\sec \theta$
Hence putting in the respective values of x and y in LHS we get:-
$LHS = {\left( {a\sec \theta + b\tan \theta } \right)^2} - {\left( {a\tan \theta + b\sec \theta } \right)^2}$
Now applying the following identity on both he terms:-
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
We get:-
$LHS = {\left( {a\sec \theta } \right)^2} + {\left( {b\tan \theta } \right)^2} + 2\left( {a\sec \theta } \right)\left( {b\tan \theta } \right) - \left[ {{{\left( {a\tan \theta } \right)}^2} + {{\left( {b\sec \theta } \right)}^2} + 2\left( {b\sec \theta } \right)\left( {a\tan \theta } \right)} \right]$
Now simplifying it further we get:-
$LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - \left[ {{a^2}{{\tan }^2}\theta + {b^2}{{\sec }^2}\theta + 2ab\sec \theta \tan \theta } \right] \\ \Rightarrow LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta + 2ab\sec \theta \tan \theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta - 2ab\sec \theta \tan \theta \\$
Now cancelling the required terms we get:-
$LHS = {a^2}{\sec ^2}\theta + {b^2}{\tan ^2}\theta - {a^2}{\tan ^2}\theta - {b^2}{\sec ^2}\theta$
Now taking ${a^2}$ and ${b^2}$ common we get:-
$LHS = {a^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right) - {b^2}\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)$
Again taking common we get:-
$LHS = \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\left( {{a^2} - {b^2}} \right)$
Now applying the following identity:-
$1 + {\tan ^2}\theta = {\sec ^2}\theta \\ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1 \\$
We get:-
$\Rightarrow LHS = \left( {{a^2} - {b^2}} \right) \\ {\text{ }} = RHS \\$
Therefore,
$LHS = RHS$
Hence proved.

Note: The student may make mistakes while applying the identity, so the identity should be first simplified in the required form and then apply it.
The student can also use the following identity on the initial stage:
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Hence,
${{\text{x}}^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)$
Then put in the values of x and y and proceed further to get the desired answer.