Answer :
Let's solve the expression step-by-step.
Given the expression:
[tex]\[ \frac{a + \sqrt{a^2 - b^2}}{a - \sqrt{a^2 - b^2}} + \frac{a - \sqrt{a^2 - b^2}}{a + \sqrt{a^2 - b^2}} \][/tex]
For simplicity, assign:
[tex]\[ x = a + \sqrt{a^2 - b^2} \][/tex]
[tex]\[ y = a - \sqrt{a^2 - b^2} \][/tex]
Rewrite the expression in terms of \(x\) and \(y\):
[tex]\[ \frac{x}{y} + \frac{y}{x} \][/tex]
We know that the sum of two fractions can be expressed as:
[tex]\[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \][/tex]
Next, find \(x^2\) and \(y^2\).
[tex]\[ x = a + \sqrt{a^2 - b^2} \implies x^2 = \left(a + \sqrt{a^2 - b^2}\right)^2 = a^2 + 2a\sqrt{a^2 - b^2} + (a^2 - b^2) = 2a^2 - b^2 + 2a\sqrt{a^2 - b^2} \][/tex]
[tex]\[ y = a - \sqrt{a^2 - b^2} \implies y^2 = \left(a - \sqrt{a^2 - b^2}\right)^2 = a^2 - 2a\sqrt{a^2 - b^2} + (a^2 - b^2) = 2a^2 - b^2 - 2a\sqrt{a^2 - b^2} \][/tex]
Sum of \(x^2\) and \(y^2\):
[tex]\[ x^2 + y^2 = \left(2a^2 - b^2 + 2a\sqrt{a^2 - b^2}\right) + \left(2a^2 - b^2 - 2a\sqrt{a^2 - b^2}\right) = 4a^2 - 2b^2 \][/tex]
Next, find \(xy\):
[tex]\[ x \cdot y = \left(a + \sqrt{a^2 - b^2}\right)\left(a - \sqrt{a^2 - b^2}\right) = a^2 - \left(\sqrt{a^2 - b^2}\right)^2 = a^2 - (a^2 - b^2) = b^2 \][/tex]
Now, use the values of \(x^2 + y^2\) and \(xy\):
[tex]\[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = \frac{4a^2 - 2b^2}{b^2} = \frac{4a^2}{b^2} - \frac{2b^2}{b^2} = \frac{4a^2}{b^2} - 2 \][/tex]
Thus, the value of the given expression is:
[tex]\[ \boxed{\frac{4a^2}{b^2} - 2} \][/tex]
Given the expression:
[tex]\[ \frac{a + \sqrt{a^2 - b^2}}{a - \sqrt{a^2 - b^2}} + \frac{a - \sqrt{a^2 - b^2}}{a + \sqrt{a^2 - b^2}} \][/tex]
For simplicity, assign:
[tex]\[ x = a + \sqrt{a^2 - b^2} \][/tex]
[tex]\[ y = a - \sqrt{a^2 - b^2} \][/tex]
Rewrite the expression in terms of \(x\) and \(y\):
[tex]\[ \frac{x}{y} + \frac{y}{x} \][/tex]
We know that the sum of two fractions can be expressed as:
[tex]\[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \][/tex]
Next, find \(x^2\) and \(y^2\).
[tex]\[ x = a + \sqrt{a^2 - b^2} \implies x^2 = \left(a + \sqrt{a^2 - b^2}\right)^2 = a^2 + 2a\sqrt{a^2 - b^2} + (a^2 - b^2) = 2a^2 - b^2 + 2a\sqrt{a^2 - b^2} \][/tex]
[tex]\[ y = a - \sqrt{a^2 - b^2} \implies y^2 = \left(a - \sqrt{a^2 - b^2}\right)^2 = a^2 - 2a\sqrt{a^2 - b^2} + (a^2 - b^2) = 2a^2 - b^2 - 2a\sqrt{a^2 - b^2} \][/tex]
Sum of \(x^2\) and \(y^2\):
[tex]\[ x^2 + y^2 = \left(2a^2 - b^2 + 2a\sqrt{a^2 - b^2}\right) + \left(2a^2 - b^2 - 2a\sqrt{a^2 - b^2}\right) = 4a^2 - 2b^2 \][/tex]
Next, find \(xy\):
[tex]\[ x \cdot y = \left(a + \sqrt{a^2 - b^2}\right)\left(a - \sqrt{a^2 - b^2}\right) = a^2 - \left(\sqrt{a^2 - b^2}\right)^2 = a^2 - (a^2 - b^2) = b^2 \][/tex]
Now, use the values of \(x^2 + y^2\) and \(xy\):
[tex]\[ \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = \frac{4a^2 - 2b^2}{b^2} = \frac{4a^2}{b^2} - \frac{2b^2}{b^2} = \frac{4a^2}{b^2} - 2 \][/tex]
Thus, the value of the given expression is:
[tex]\[ \boxed{\frac{4a^2}{b^2} - 2} \][/tex]