গণিত প্রশ্ন

উত্তর ও ব্যাখ্যা

1. ক)

   $\sec^2\theta-\tan^2\theta=1$ বা, $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$ বা, $\frac23(\sec\theta-\tan\theta)=1$ বা, $\sec\theta-\tan\theta=\frac32$

   $\therefore\tan\theta-\sec\theta=\mathbf{-\frac32}$

2. খ)

   $\frac1{2-p^2}=\frac1{2-\cos^2\theta}=\frac1{1+\sin^2\theta}$

   $\frac1{2+q^2}=\frac1{2+\cot^2\theta}=\frac1{1+\text{cosec}^2\theta}=\frac1{1+\frac1{\sin^2\theta}}=\frac{\sin^2\theta}{\sin^2\theta+1}$

   যোগ করে, $\frac{1+\sin^2\theta}{1+\sin^2\theta}=1$ (প্রমাণিত)

3. গ)

   $\frac{p-q}{p+q}=\frac{\cos\theta-\cot\theta}{\cos\theta+\cot\theta}=\frac{\cos\theta\left(1-\frac1{\sin\theta}\right)}{\cos\theta\left(1+\frac1{\sin\theta}\right)}=\frac{\sin\theta-1}{\sin\theta+1}$

   শর্তমতে, $\frac{\sin\theta-1}{\sin\theta+1}=\frac{\sqrt3-2}{\sqrt3+2}$

   যোজন-বিয়োজন করে, $\frac{(\sin\theta-1)+(\sin\theta+1)}{(\sin\theta+1)-(\sin\theta-1)}=\frac{(\sqrt3-2)+(\sqrt3+2)}{(\sqrt3+2)-(\sqrt3-2)}$ বা, $\frac{2\sin\theta}{2}=\frac{2\sqrt3}4$

   বা, $\sin\theta=\frac{\sqrt3}2=\sin60^\circ$

   $\therefore\theta=\mathbf{60^\circ}$