$\frac{\sin \theta }{1 - \cos \theta }= m, \frac{1 - \cos \theta }{\sin \theta } = n$ এবং p = cot $\theta$

দেওয়া আছে,  $\tan A = \frac{a}{b}$

আমরা জানি, $se{c}^{2}A=1+ta{n}^{2}A$

$=1+{\left(\frac{a}{b}\right)}^2=1+\frac{a^2}{b^2}=\frac{b^2+a^2}{b^2}=\frac{a^2+b^2}{b^2}$

নির্ণেয় মান : $\frac{a^2+b^2}{b^2}$

দেওয়া আছে, $m = \frac{\sin \theta }{1 - \cos \theta }, n= \frac{1-\cos \theta }{\sin \theta }$ এবং $p=cot \theta$

বামপক্ষ = m - n

$=\frac{\sin \theta }{1-cos \theta }-\frac{1-cos \theta }{\sin \theta }= \frac{{\sin }^2 \theta - (1-\cos \theta {)}^2}{sin \theta (1-cos \theta )}$

$=\frac{{\sin }^2 \theta - \left(1-2\cos \theta +{\cos }^2 \theta \right)}{sin \theta (1-cos \theta )}$

$=\frac{ {\sin }^2 \theta -1+2 \cos \theta -{\cos }^2}{sin \theta (1-cos \theta )}$

$=\frac{{\sin }^2 \theta +{\cos }^2 \theta -1+2\cos \theta -2{\cos }^2 \theta }{\sin \theta \left(1+\cos \theta \right)}$

$=\frac{1 - 1 + 2 \cos \theta \left(1 - \cos \theta \right)}{\sin \theta (1 - \cos \theta )}$

$= \frac{2 \cos \theta (1 - \cos \theta )}{\sin \theta (1 - \cos \theta )}=2. \frac{\cos \theta }{\sin \theta }=2.cot \theta =2p$

= ডানপক্ষ

$\therefore$ m - n = 2p (প্রমানিত)

দেওয়া আছে,  m = 3p

বা, $\frac{\sin \theta }{1 - \cos \theta }=3 cot \theta$

বা, $\frac{\sin \theta }{1 - \cos \theta }= 3$ .$\frac{\cos \theta }{\sin \theta }$

বা,  $si{n}^2 \theta =3cos \theta -3co{s}^2 \theta$

বা, $1-co{s}^2 \theta -3cos \theta +3co{s}^2 \theta =0$

বা, $2co{s}^2 \theta -3 cos \theta +1=0$

বা, $2co{s}^2 \theta -2cos \theta -cos \theta +1=0$

বা, $2\cos \theta (\cos \theta - 1) - (\cos \theta - 1) = 0$

বা, $(\cos \theta - 1) (2\cos \theta - 1) = 0$

হয়, $\cos \theta - 1 = 0$           |  অথবা, $2\cos \theta - 1 =0$

বা,  $\cos \theta = 1$                          বা, $2\cos \theta = 1$

বা,  cos $\theta$ = cos 0°                   বা,   $\cos \theta = \frac{1}{2}$

$\therefore$ $\theta$ = 0°                                  বা, cos $\theta$ = cos 60°

                                                       $\therefore$   $\theta$ = 60°

কিন্তু $\theta \ne$0° কারণ সূক্ষ্মকোণ।

সুতরাং $\theta$ = 60°

নির্ণেয় মান : $\theta$ = 60°