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I tried dividing both sides by $\cos$ and then squaring them. Then I converted $\sec^2(x)$ to $\tan^2(x)$ but still no luck even after substituting values.

2 Answers2

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hint...Put $t=\tan\frac 12\theta$ so that $\cos\theta=\frac{1-t^2}{1+t^2}$ and $\sin\theta=\frac{2t}{1+t^2}$.

Then consider the sum and product of the roots of the quadratic in $t$

David Quinn
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$$p(\cos\beta-\cos\alpha)=q(\sin\alpha-\sin\beta)$$

Now apply this