Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
If $(x-a)\cos\theta + y\sin\theta = (x-a)\cos\phi + y\sin\phi = a$, $\tan\frac{\theta}{2} - \tan\frac{\phi}{2} = 2e$ and $\theta, \phi$ are unequal angles less than $360°$, then $y^2$ is equal to:
$2ax - (1+e^2)x^2$
$2ax - (1-e^2)x^2$
$2ax + (1-e^2)x^2$
$2ax + (1+e^2)x^2$
Step-by-Step Solution
Key Concept: Convert trigonometric expressions using half-angle substitution formulas to transform the problem into finding roots of a quadratic equation.
Using the half-angle formulas $\cos\theta = \frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$ and $\sin\theta = \frac{2\tan(\theta/2)}{1+\tan^2(\theta/2)}$, we set $\tan(\theta/2)$ and $\tan(\phi/2)$ as roots of the quadratic $(x-a)\frac{1-t^2}{1+t^2} + x \cdot \frac{2t}{1+t^2} = a$. Simplifying gives $ux^2 - 2yx + 2a - x = 0$, or $ux^2 - 2yu - x = 0$. Therefore $\tan(\theta/2) \cdot \tan(\phi/2) = \frac{2u - x}{x}$.
Correct Answer: 2