Question

If $x=\sqrt{a} \tan \alpha$ and $y=\sqrt{a} \sec \alpha$, find the value of $y^{2}-x^{2}$.

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Answer

To solve for $y^2 - x^2$, let's use the given expressions for ( x ) and ( y ):

[ x = \sqrt{a} \tan \alpha ] [ y = \sqrt{a} \sec \alpha ]

First, we square both expressions:

[ x^2 = (\sqrt{a} \tan \alpha)^2 = a \tan^2 \alpha ] [ y^2 = (\sqrt{a} \sec \alpha)^2 = a \sec^2 \alpha ]

Now substitute these squared terms into the expression ( y^2 - x^2 ):

[ y^2 - x^2 = a \sec^2 \alpha - a \tan^2 \alpha ]

Factor out ( a ) from the expression:

[ y^2 - x^2 = a (\sec^2 \alpha - \tan^2 \alpha) ]

Utilize the trigonometric identity for ( \sec^2 \alpha - \tan^2 \alpha ):

[ \sec^2 \alpha - \tan^2 \alpha = 1 ]

So the expression simplifies to:

[ y^2 - x^2 = a \cdot 1 = a ]

Therefore, the value of ( y^2 - x^2 ) is:

[ \boxed{a} ]


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