$\frac{1}{A}=\frac{1 - sec \theta + \tan \theta }{sec \theta + \tan \theta - 1}$, $B =\frac{1 + \sin \theta }{\cos \theta }$

দেওয়া আছে,

$\theta =\frac{{\mathrm{\pi }}^c}{3}={\left(\frac{\mathrm{\pi }}{3}\times \frac{180}{\mathrm{\pi }}\right)}^{°}$

$\therefore$ $\theta = 60°$

এবং $B=\frac{1+\sin \theta }{\cos \theta }=\frac{1+\sin 60°}{\cos 60°}=\frac{1+{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{{\displaystyle \frac{2+\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{2+\sqrt{3}}{2}\times 2 = 2+\sqrt{3}$

নির্ণেয় B এর মান :  $2+\sqrt{3}$

দেওয়া আছে, $\frac{1}{A}=\frac{1 - sec \theta + \tan \theta }{sec \theta + \tan \theta - 1}$

বা, $A=\frac{\tan \theta + sec \theta - 1}{\tan \theta - sec \theta + 1}$

= $\frac{\tan \theta + sec \theta \left(se{c}^2 \theta -{\tan }^2\theta \right)}{\tan \theta - sec \theta + 1}$ $\left[\because se{c}^2\theta -{\tan }^2 \theta -1\right]$

$= \frac{(\tan \theta + sec \theta ) - (sec \theta + \tan \theta )(sec \theta - \tan \theta )}{(\tan \theta - sec \theta + 1)}$

$= \frac{(\tan \theta + sec \theta )(1 - sec \theta + \tan \theta )}{(\tan \theta - sec \theta + 1)}$

$=\frac{(\tan \theta + sec \theta ) (\tan \theta - sec \theta + 1)}{(\tan - sec + 1) }= \tan \theta + sec \theta .$

এবং $B=\frac{1 + \sin \theta }{\cos \theta }= \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }= sec \theta + \tan \theta$

বামপক্ষ = A - B = $\tan \theta +sec\theta$ - $sec \theta - \tan \theta$ = 0 = ডানপক্ষ

$\therefore$ A - B = 0 (দেখানো হলো)

দেওয়া আছে, $B=\frac{1 + \sin \theta }{\cos \theta }$ এবং $B = \sqrt{3}$

বা, $\frac{1 + \sin \theta }{\cos \theta }=\sqrt{3}$

বা, $1 + \sin \theta =\sqrt{3} \cos \theta$

বা, $(1+sin \theta {)}^2={\left(\sqrt{3}\cos \theta \right)}^2$ [বর্গ করে]

বা, $1+2sin \theta +si{n}^2\theta =3co{s}^2\theta$

বা, $1+2 \sin \theta + {\sin }^2 \theta = 3(1 - {\sin }^2 \theta )$

বা, $1+2sin\theta +si{n}^2\theta =3-3si{n}^2\theta$

বা, $1+2sin \theta +si{n}^2\theta -3+3si{n}^2\theta =0$

বা, $4si{n}^2\theta +2sin \theta -2=0$

বা, $2si{n}^2\theta +\sin \theta -1=0$

বা, $2si{n}^2\theta +2sin \theta -sin \theta -1=0$

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

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

হয়, $\sin \theta + 1 = 0$

বা, $\sin \theta = - 1$

$\sin \theta = - 1$ হলে,

$\sin \theta = - 1 = \sin 270°$

$\therefore \theta = 270°$

অথবা, $2\sin \theta - 1 = 0$

বা, $\sin \theta = \frac{1}{2}$

আবার, $\sin \theta = \frac{1}{2}$ হলে, প্রথম চতুর্ভাগে, $\sin \theta = \frac{1}{2}=\sin 30°$

$\therefore \theta = 30°$

এবং দ্বিতীয় চতুর্ভাগে, $\sin \theta =\frac{1}{2}= \sin (180° - 30°) = \sin 150°$

$\therefore \theta = 150°$

কিন্তু $\theta = 150°, 270°$ প্রদত্ত সমীকরণকে সিদ্ধ করে না।

$\therefore \theta = 150°, 270°$ গ্রহণযোগ্য নয়।

নির্ণেয় মান : $\theta = 30°$