Answer

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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.

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.

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