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  4. If 2y cos θ = x sin θ and 2x sec θ - y cosec θ = 3, then x2 + 4y2 =

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Q. If $2y \,cos\,\theta = x \,sin \,\theta\, and\, 2x \,sec\,\theta - y\, cosec\,\theta = 3,$ then $x^2 + 4y^2 =$

Trigonometric Functions

Solution:

Given that $2y \,cos\,\theta = x \,sin \,\theta\quad ...\left(i\right)$
and $2x \,sec\,\theta - y\, cosec\,\theta = 3\quad ...\left(ii\right)$
$\Rightarrow \frac{2x}{cos\,\theta } - \frac{y}{sin \,\theta } = 3$
$\Rightarrow 2x\,sin\,\theta - y \,cos\,\theta -3sin\,\theta \,cos\,\theta = 0 \quad...\left(iii\right)$
Solving $\left(i\right)$ and $\left(iii\right)$, we get $y = sin\,\theta$ and $x = 2 \,cos\,\theta $
Now, $x^{2} + 4y^{2} = 4cos^{2} \,\theta + 4sin^{2} \,\theta$
$= 4\left(cos^{2} \,\theta + sin^{2} \,\theta \right) = 4$