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Question 142
If $$ 0 \leq \theta \leq \frac{\pi}{2}$$, $$2ycos\theta=sin\theta$$ and $$\frac{x}{2cosec\theta}=y$$, then the value of $$x^2-4y^2$$Â is
Solution
$$2y=tan\theta$$
$$x=2ycosec\theta$$
Hence value of $$x^2 - 4y^2 $$ = $$4y^2(cosec^2\theta - 1)$$
or $$tan^2\theta cot^2\theta$$ = 1
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