$cot \theta + cosec \theta = m$ একটি ত্রিকোণমিতিক সমীকরণ।

দেওয়া আছে,  $cot \theta + \cos ec \theta = m$

$m =\frac{5}{4}$ হলে,

$\therefore$ $\cos ec \theta + cot \theta = \frac{5}{4}$

আমরা জানি, $cose{c}^2\theta -co{t}^2\theta =1$

বা, $(cosec \theta + cot \theta )(cosec \theta - cot \theta ) = 1$

বা, $\frac{5}{4} (cosec \theta - cot \theta ) = 1$

বা, $cosec \theta - cot \theta = \frac{4}{5}$

বা, $- (cot \theta - cosec \theta ) = \frac{4}{5}$

$\therefore$ $cot \theta - cosec \theta = - \frac{4}{5}$

নির্ণেয় মান : $- \frac{4}{5}$

দেওয়া আছে, $cot \theta + cosec \theta = m$

এখন, m = 2 হলে, $\cos ec \theta + cot \theta = 2$ _________ (i)

আমরা জানি, $cose{c}^2\theta -co{t}^2\theta =1$

বা, $(\cos ec \theta + cot \theta )(cosec \theta - cot \theta ) = 1$

বা, $2(cosec \theta - cot \theta ) = 1$

$\therefore$ $\cos ec \theta - cot \theta = \frac{1}{2}$ _________ (ii)

সমীকরণ (i) ও (ii) যোগ করে পাই,

$2 \cos ec \theta = 2 + \frac{1}{2}=\frac{5}{2}$

বা, $\cos ec \theta =\frac{5}{4}$

বা, $\frac{1}{\sin \theta }=\frac{5}{4}$

$\therefore$ $\sin \theta =\frac{4}{5}$

আবার, সমীকরণ (i) নং হতে (ii) নং বিয়োগ করে পাই,

$2 cot \theta = 2 - \frac{1}{2}= \frac{3}{2}$

বা, $cot \theta = \frac{3}{4}$

বা, $\frac{\cos \theta }{\sin \theta }=\frac{3}{4}$

বা, $\cos \theta = \frac{3}{4}\times \sin \theta$

$\therefore$ $\cos \theta = \frac{3}{4}\times \frac{4}{5}= \frac{3}{5}$

বামপক্ষ $= \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$

$=\frac{{\displaystyle \frac{4}{5}}- {\displaystyle \frac{3}{5}}+ 1}{{\displaystyle \frac{4}{5}}+{\displaystyle \frac{3}{5}}- 1}= \frac{{\displaystyle \frac{4 - 3 + 5}{5}}}{{\displaystyle \frac{4+3-5}{5}}} = \frac{6}{5}\times \frac{5}{2}=3$

ডানপক্ষ $=\frac{2\cos \theta }{1 - \cos \theta }=\frac{2\times {\displaystyle \frac{ 3}{5}}}{1-{\displaystyle \frac{3}{5}}}= \frac{{\displaystyle \frac{6}{5}}}{{\displaystyle \frac{5 - 3}{5}}}= \frac{6}{5}\times \frac{5}{2} = 3$

$\therefore$ $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}$= $\frac{2\cos \theta }{1 - \cos \theta }$ (দেখানো হলো)

দেওয়া আছে, $cot \theta + \cos ec \theta = m$

এখন, $m= -\sqrt{3}$ হলে, $cot \theta +\cos ec \theta =-\sqrt{3}$

বা $\cos ec \theta + cot =-\sqrt{3}$ _______ (i)

আমরা জানি,

$cose{c}^2\theta -co{t}^2\theta =1$

বা, $(\cos ec \theta + cot \theta )(cosec \theta - cot \theta ) = 1$

বা, $- \sqrt{3} (cosec \theta - cot \theta ) = 1$

$\therefore$ $\cos ec \theta - cot \theta = - \frac{1}{\sqrt{3}}$ _______ (ii)

(i) নং থেকে (ii) নং বিয়োগ করে পাই,

$2cot \theta = - \sqrt{3} + \frac{1}{\sqrt{3}}=\frac{- 3 + 1}{\sqrt{3}}$

বা, $cot \theta = - \frac{2}{\sqrt{3}}$

বা, $\frac{1}{\tan \theta }= \frac{-2}{2\sqrt{3}}= -\frac{1}{\sqrt{3}}$

বা, $\tan \theta =-\sqrt{3}= - \tan \frac{\mathrm{\pi }}{3}$

বা, $\tan \theta = \tan \left(-\frac{\mathrm{\pi }}{3}\right)= \tan \left(\pi -\frac{\mathrm{\pi }}{3}\right)$

$\therefore$  $\theta = - \frac{\mathrm{\pi }}{3}, \frac{2\mathrm{\pi }}{3}$

নির্ণেয় মান : $\theta = - \frac{\mathrm{\pi }}{3}, \frac{2\mathrm{\pi }}{3}$