Question

If \( \sin \theta = \frac{b}{a} \), then \( \frac{\sqrt{a+b}} {\sqrt{a-b}} + \frac{\sqrt{a-b}} {\sqrt{a+b}} \) is equal to:

• A. \( \frac{2}{\cos \theta} \)
• B. \( \frac{1}{\cos \theta} \)
• C. \( \frac{2}{\sqrt{\cos \theta}} \)
• D. \( \frac{1}{\cos \theta} \)
• E. 1

Hint:

Utilize algebraic identities and trigonometric relations to simplify complex expressions, converting terms into forms that can leverage known identities.

Answer 1

The Correct Option is A

Given that \( \sin \theta = \frac{b}{a} \), we first recognize that \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{b}{a}\right)^2} = \sqrt{\frac{a^2 - b^2}{a^2}} = \frac{\sqrt{a^2 - b^2}}{a} \). 
Now consider the expression: \[ \frac{\sqrt{a+b}} {\sqrt{a-b}} + \frac{\sqrt{a-b}} {\sqrt{a+b}} \] which can be simplified by letting \( x = \sqrt{a+b} \) and \( y = \sqrt{a-b} \), thus: \[ \frac{x}{y} + \frac{y}{x} \] Using the identity \( \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} \), and substituting back: \[ \frac{(\sqrt{a+b})^2 + (\sqrt{a-b})^2}{\sqrt{a+b} \sqrt{a-b}} = \frac{a+b + a-b}{\sqrt{(a+b)(a-b)}} = \frac{2a}{\sqrt{a^2 - b^2}} \] From the earlier substitution for \( \cos \theta \), \( \sqrt{a^2 - b^2} = a \cos \theta \): \[ \frac{2a}{a \cos \theta} = \frac{2}{\cos \theta} \] Thus, the expression simplifies to \( \frac{2}{\cos \theta} \).