গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $pq = (\cot\theta + \cos\theta)(\cot\theta - \cos\theta) = \cot^2\theta - \cos^2\theta = \dfrac{\cos^2\theta}{\sin^2\theta} - \cos^2\theta = \cos^2\theta\left(\dfrac{1-\sin^2\theta}{\sin^2\theta}\right) = \cos^2\theta\cdot\dfrac{\cos^2\theta}{\sin^2\theta} = \cos^2\theta\cot^2\theta$।
   অতএব, $pq = \cot^2\theta\cdot\cos^2\theta$। (দেখানো হলো)

2. খ)

   $p^2 - q^2 = (p+q)(p-q) = (2\cot\theta)(2\cos\theta) = 4\cot\theta\cos\theta$।
   আবার (ক) থেকে $pq = \cot^2\theta\cos^2\theta$, তাই $4\sqrt{pq} = 4\sqrt{\cot^2\theta\cos^2\theta} = 4\cot\theta\cos\theta$।
   সুতরাং, $p^2 - q^2 = 4\sqrt{pq}$। (প্রমাণিত)

3. গ)

   $\dfrac{p}{q} = \dfrac{\cot\theta+\cos\theta}{\cot\theta-\cos\theta} = \dfrac{\frac{\cos\theta}{\sin\theta}+\cos\theta}{\frac{\cos\theta}{\sin\theta}-\cos\theta} = \dfrac{1+\sin\theta}{1-\sin\theta}$
   শর্তমতে $\dfrac{1+\sin\theta}{1-\sin\theta} = \dfrac{2+\sqrt{3}}{2-\sqrt{3}}$
   যোজন-বিয়োজন করে $\dfrac{(1+\sin\theta)+(1-\sin\theta)}{(1+\sin\theta)-(1-\sin\theta)} = \dfrac{(2+\sqrt{3})+(2-\sqrt{3})}{(2+\sqrt{3})-(2-\sqrt{3})}$
   বা $\dfrac{2}{2\sin\theta} = \dfrac{4}{2\sqrt{3}}$ বা $\dfrac{1}{\sin\theta} = \dfrac{2}{\sqrt{3}}$
   তাই $\sin\theta = \dfrac{\sqrt{3}}{2} = \sin 60^\circ$।
   অতএব, $\theta = \mathbf{60^\circ}$।