Question

If \( \tan \theta = \frac{a}{b} \) then \( \frac{a\sin\theta + b\cos\theta}{a\sin\theta - b\cos\theta} = \)

• A. \( \frac{a^2+b^2}{a^2-b^2} \)
• B. \( \frac{a^2-b^2}{a^2+b^2} \)
• C. \( \frac{a+b}{a-b} \)
• D. \( \frac{a-b}{a+b} \)

Hint:

Given an expression with \(\sin\theta\) and \(\cos\theta\), and knowing \(\tan\theta\): A quick method is to divide the numerator and denominator by \(\cos\theta\). This converts \(\sin\theta\) to \(\tan\theta\) and \(\cos\theta\) to 1. Expression: \( \frac{a\sin\theta + b\cos\theta}{a\sin\theta - b\cos\theta} \) Divide by \(\cos\theta\): \( \frac{a(\sin\theta/\cos\theta) + b(\cos\theta/\cos\theta)}{a(\sin\theta/\cos\theta) - b(\cos\theta/\cos\theta)} = \frac{a\tan\theta + b}{a\tan\theta - b} \). Substitute \(\tan\theta = a/b\): \( \frac{a(a/b) + b}{a(a/b) - b} = \frac{a^2/b + b}{a^2/b - b} = \frac{(a^2+b^2)/b}{(a^2-b^2)/b} = \frac{a^2+b^2}{a^2-b^2} \).

Answer 1

The Correct Option is A

Concept: This problem can be solved by expressing \(\sin\theta\) and \(\cos\theta\) in terms of \(a\) and \(b\) using a right-angled triangle, or by dividing the numerator and denominator of the expression by \(\cos\theta\). Method 1: Dividing by \(\cos\theta\) Given expression: \( \frac{a\sin\theta + b\cos\theta}{a\sin\theta - b\cos\theta} \) Divide both the numerator and the denominator by \(\cos\theta\) (assuming \(\cos\theta \neq 0\)): Numerator: \( \frac{a\sin\theta + b\cos\theta}{\cos\theta} = a\frac{\sin\theta}{\cos\theta} + b\frac{\cos\theta}{\cos\theta} = a\tan\theta + b \) Denominator: \( \frac{a\sin\theta - b\cos\theta}{\cos\theta} = a\frac{\sin\theta}{\cos\theta} - b\frac{\cos\theta}{\cos\theta} = a\tan\theta - b \) So the expression becomes: \[ \frac{a\tan\theta + b}{a\tan\theta - b} \] We are given \( \tan\theta = \frac{a}{b} \). Substitute this into the expression: \[ \frac{a\left(\frac{a}{b}\right) + b}{a\left(\frac{a}{b}\right) - b} = \frac{\frac{a^2}{b} + b}{\frac{a^2}{b} - b} \] To simplify the complex fraction, find a common denominator (\(b\)) for the terms in the numerator and denominator: Numerator: \( \frac{a^2}{b} + \frac{b^2}{b} = \frac{a^2+b^2}{b} \) Denominator: \( \frac{a^2}{b} - \frac{b^2}{b} = \frac{a^2-b^2}{b} \) So the expression is: \[ \frac{\left(\frac{a^2+b^2}{b}\right)}{\left(\frac{a^2-b^2}{b}\right)} = \frac{a^2+b^2}{b} \times \frac{b}{a^2-b^2} = \frac{a^2+b^2}{a^2-b^2} \] Method 2: Using a right-angled triangle If \( \tan\theta = \frac{a}{b} \), we can consider a right-angled triangle where: Opposite side = \(a\) Adjacent side = \(b\) Hypotenuse \(h = \sqrt{a^2+b^2}\) Then, \(\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{a}{\sqrt{a^2+b^2}}\) And \(\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{b}{\sqrt{a^2+b^2}}\) Substitute these into the expression \( \frac{a\sin\theta + b\cos\theta}{a\sin\theta - b\cos\theta} \): Numerator: \( a\left(\frac{a}{\sqrt{a^2+b^2}}\right) + b\left(\frac{b}{\sqrt{a^2+b^2}}\right) = \frac{a^2}{\sqrt{a^2+b^2}} + \frac{b^2}{\sqrt{a^2+b^2}} = \frac{a^2+b^2}{\sqrt{a^2+b^2}} \) Denominator: \( a\left(\frac{a}{\sqrt{a^2+b^2}}\right) - b\left(\frac{b}{\sqrt{a^2+b^2}}\right) = \frac{a^2}{\sqrt{a^2+b^2}} - \frac{b^2}{\sqrt{a^2+b^2}} = \frac{a^2-b^2}{\sqrt{a^2+b^2}} \) The expression becomes: \[ \frac{\left(\frac{a^2+b^2}{\sqrt{a^2+b^2}}\right)}{\left(\frac{a^2-b^2}{\sqrt{a^2+b^2}}\right)} = \frac{a^2+b^2}{a^2-b^2} \] Both methods yield \( \frac{a^2+b^2}{a^2-b^2} \).