If tanθ = \frac{a}{b} then \frac{\left(\right. asinθ - bcosθ \left.\right)}{\left(\right. asinθ + bcosθ \left.\right)} = ?

Foundation

Mathematics Foundation

Introduction to Trigonometry

Question

If $tanθ = \frac{a}{b}$ then $\frac{(asinθ - bcosθ)}{(asinθ + bcosθ)} = ?$

$\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}$

$\frac{b^{2}}{(a^{2} + b^{2})}$

$\frac{a^{2}}{(a^{2} + b^{2})}$

$\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$

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Given tanθ = a/b, we can represent sinθ and cosθ in terms of a and b. Let sinθ = a/√(a²+b²) and cosθ = b/√(a²+b²).

Substitute into the expression:

$\frac{(a sinθ - b cosθ)}{(a sinθ + b cosθ)} = \frac{a \cdot \frac{a}{\sqrt{a^{2} + b^{2}}} - b \cdot \frac{b}{\sqrt{a^{2} + b^{2}}}}{a \cdot \frac{a}{\sqrt{a^{2} + b^{2}}} + b \cdot \frac{b}{\sqrt{a^{2} + b^{2}}}} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Final Answer: $\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$