$\sin \theta =p. \cos \theta = q . \tan \theta = r$ যেখানে $\theta$ সূক্ষ্মকোণ।

দেওয়া আছে, tan $\theta$ = r

এখন, $r=\sqrt{{\left(3\right)}^{-1}}$ হলে,

$\tan \theta =\sqrt{{\left(3\right)}^{-1}}$

বা, $\tan \theta =\frac{1}{3}= \frac{1}{\sqrt{3}}= \tan 30°$

$\therefore$ $\theta$ = 30°

নির্ণেয় মান: $\theta$ = 30°

দেওয়া আছে, sin $\theta$ = p এবং cos $\theta$ = q

$p+q=\sqrt{2}$ হলে

$\sin \theta + \cos \theta = \sqrt{2}$

বা, $\sin \theta = \sqrt{2} - \cos \theta$

বা, ${\sin }^2 \theta = {\left(\sqrt{2} - \cos \theta \right)}^2$

বা, $si{n}^2 \theta ={\left(\sqrt{2}\right)}^2-2.\sqrt{2}.cos \theta +co{s}^2 \theta$

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

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

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

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

বা, ${\left(2 \cos \theta -1\right)}^2=0$

বা, $\sqrt{2} \cos \theta - 1 = 0$

বা, $\sqrt{2} \cos \theta = 1$

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

বা, cos $\theta$ = cos 45°

$\therefore$ $\theta$ = 45°(প্রমাণিত)

দেওয়া আছে sin $\theta$ = p এবং cos $\theta$ = q

$7p^2+3q^2=4$ হলে

$7si{n}^2 \theta +3co{s}^2 \theta =4$

বা, $7si{n}^2 \theta +3 (1-si{n}^2 \theta )=4$

বা, $7si{n}^2 \theta +3-3si{n}^2 \theta =4$

বা, $4si{n}^2 \theta +3=4$

বা, $4si{n}^2 \theta =4-3$

বা, $4si{n}^2 \theta =1$

বা, $si{n}^2 \theta =\frac{1}{4}$

বা, $\sin \theta = \sqrt{\frac{1}{4}}= \frac{1}{2}= \sin 30°$

$\therefore$ $\theta$ = 30°

বামপক্ষ = $\tan \theta = \tan 30°$ = $\frac{1}{\sqrt{3}}$ = ডানপক্ষ

$\therefore$ $\tan \theta = \frac{1}{\sqrt{3}}$ (দেখানো হলো)