Question

# Prove the following identities:If $x=a\sec \theta +b\tan \theta$ and $y=a\tan \theta +b\sec \theta$, prove that ${{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}$

Hint: First of all, find the expression for ${{x}^{2}}$ and ${{y}^{2}}$ by using the formula ${{\left( p+q\right)}^{2}}={{p}^{2}}+{{q}^{2}}+2pq$. Then take the difference that ${{x}^{2}}-{{y}^{2}}$. Then use identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ to prove the desired result.

Complete step by step solution:
We are given that $x=a\sec \theta +b\tan \theta$ and $y=a\tan \theta +b\sec \theta$, we have to
prove that ${{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}$
Let us first consider the expression for x given in the question.
$x=asec\theta +b\tan \theta$
By squaring both sides of the above equation, we get,
${{x}^{2}}={{\left( a\sec \theta +b\tan \theta \right)}^{2}}$
We know that ${{\left( p+q \right)}^{2}}={{p}^{2}}+{{q}^{2}}+2pq$. By applying this formula in RHS of the
above equation by considering $p=a\sec \theta$ and $q=b\tan \theta$, we get,
${{x}^{2}}={{\left( a\sec \theta \right)}^{2}}+{{\left( b\tan \theta \right)}^{2}}+2\left( a\sec \theta \right)\left( b\tan \theta \right)$
We can also write the above equation as
${{x}^{2}}={{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta +2ab\sec \theta \tan \theta ....\left( i \right)$
Now, let us consider the expression for y given in the question, we get,
$y=a\tan \theta +b\sec \theta$
By squaring both sides of the above equation, we get,
${{y}^{2}}={{\left( a\tan \theta +b\sec \theta \right)}^{2}}$
We know that ${{\left( p+q \right)}^{2}}={{p}^{2}}+{{q}^{2}}+2pq$. By applying this formula in RHS of the
above equation by considering $p=a\tan \theta$ and $q=b\sec \theta$, we get
${{y}^{2}}={{\left( a\tan \theta \right)}^{2}}+{{\left( b\sec \theta \right)}^{2}}+2\left( a\tan \theta \right)\left( b\sec \theta \right)$
We can also write the above expression as,

${{y}^{2}}={{a}^{2}}{{\tan }^{2}}\theta +{{b}^{2}}{{\sec }^{2}}\theta +2ab\sec \theta \tan \theta ....\left( ii \right)$
Now, by subtracting equation (ii) from (i), we get
${{x}^{2}}-{{y}^{2}}=\left( {{a}^{2}}{{\sec }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta +2ab\sec \theta \tan \theta \right)-\left( {{a}^{2}}{{\tan }^{2}}\theta +{{b}^{2}}{{\sec }^{2}}\theta +2ab\sec \theta \tan \theta \right)$
By rearranging the terms of the above equation, we get,
${{x}^{2}}-{{y}^{2}}={{a}^{2}}{{\sec }^{2}}\theta -{{a}^{2}}{{\tan }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta - {{b}^{2}}{{\sec }^{2}}\theta +2ab\sec \theta \tan \theta -2ab\sec \theta \tan \theta$
By canceling the like terms in the above equation, we get,
${{x}^{2}}-{{y}^{2}}={{a}^{2}}{{\sec }^{2}}\theta -{{a}^{2}}{{\tan }^{2}}\theta +{{b}^{2}}{{\tan }^{2}}\theta - {{b}^{2}}{{\sec }^{2}}\theta$
By taking ${{a}^{2}}$ and ${{b}^{2}}$common, we can write above the equation as,
${{x}^{2}}-{{y}^{2}}={{a}^{2}}\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)+{{b}^{2}}\left( {{\tan }^{2}}\theta -{{\sec }^{2}}\theta \right)$
We know that ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$ or ${{\tan }^{2}}\theta -{{\sec }^{2}}\theta =- 1$.
By substituting these in the above equation, we get,
${{x}^{2}}-{{y}^{2}}={{a}^{2}}\left( 1 \right)+{{b}^{2}}\left( -1 \right)$
Or, ${{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}$
Hence proved.

Note: Here students must note that they must subtract ${{y}^{2}}$ from ${{x}^{2}}$. Students often make the mistake of subtracting expression of ${{x}^{2}}$ from the expression of ${{y}^{2}}$ and writing the same in RHS and equating it with $\left( {{x}^{2}}-{{y}^{2}} \right)$ in LHS while what they calculated is $\left( {{y}^{2}}-{{x}^{2}} \right)$. So this mistake must be avoided.