If sin alpha = p,then the quadratic equation | vedclass

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Question

(A) Given $\sin \alpha = p$. We know that $\sin \alpha = \frac{2 	an \frac{\alpha}{2}}{1 + 	an^2 \frac{\alpha}{2}} = p$. Let $t = 	an \frac{\alpha}

Vedclass · Accepted Answer

$p x^2 - 2x + p = 0$

Vedclass · Answer

(A) Given $\sin \alpha = p$. We know that $\sin \alpha = \frac{2 	an \frac{\alpha}{2}}{1 + 	an^2 \frac{\alpha}{2}} = p$. Let $t = 	an \frac{\alpha}{2}$. Then $\frac{2t}{1 + t^2} = p$,which implies $2t = p + pt^2$,or $pt^2 - 2t + p = 0$. Dividing by $p$,we get $t^2 - \frac{2}{p} t + 1 = 0$. The roots of this equation are $t_1 = 	an \frac{\alpha}{2}$ and $t_2 = \frac{1}{	an \frac{\alpha}{2}} = \cot \frac{\alpha}{2}$. The sum of roots is $	an \frac{\alpha}{2} + \cot \frac{\alpha}{2} = \frac{2}{p}$ and the product of roots is $	an \frac{\alpha}{2} 	imes \cot \frac{\alpha}{2} = 1$. The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Substituting the values,we get $x^2 - \frac{2}{p} x + 1 = 0$,which simplifies to $px^2 - 2x + p = 0$. Thus,option $(A)$ is correct.

If $\sin \alpha = p$,then the quadratic equation whose roots are $\tan \frac{\alpha}{2}$ and $\cot \frac{\alpha}{2}$ is

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