Question

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

Answer 1

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.