সৃজনশীলঃ ‘3cot2⁡θ+cosec2θ=P এবং Q=−sin⁡(−θ)+cos⁡(−θ)sec⁡(−θ)+tan⁡(−θ)…’ ঢাকা বোর্ড ২০২৪ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে,

$3{\cot }^2\theta +{\mathrm{cosec}}^2\theta =P$

এখন, $P=5$ হলে, $3{\cot }^2\theta +{\mathrm{cosec}}^2\theta =5$

বা, $3({\mathrm{cosec}}^2\theta -1)+{\mathrm{cosec}}^2\theta =5[\because {\cot }^2\theta ={\mathrm{cosec}}^2\theta -1]$

বা, $3{\mathrm{cosec}}^2\theta -3+{\mathrm{cosec}}^2\theta =5$

বা, $4{\mathrm{cosec}}^2\theta =5+3=8$

বা, ${\mathrm{cosec}}^2\theta =\frac{8}{4}$

বা, $\frac{1}{{\sin }^2\theta }=2[\because \mathrm{cosec}\theta =\frac{1}{\sin \theta }]$

বা, ${\sin }^2\theta =\frac{1}{2}$

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

$\sin \theta =\frac{1}{\sqrt{2}}$ হলে, প্রথম চতুর্ভাগে, $\sin \theta =\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}\therefore \theta =\frac{\pi }{4}$

$\sin \theta =\frac{1}{\sqrt{2}}$ হলে, দ্বিতীয় চতুর্ভাগে, $\sin \theta =\sin (\pi -\frac{\pi }{4})=\sin \frac{3\pi }{4}\therefore \theta =\frac{3\pi }{4}$

আবার, $\sin \theta =-\frac{1}{\sqrt{2}}$ হলে,

তৃতীয় চতুর্ভাগে, $\sin \theta =-\frac{1}{\sqrt{2}}=-\sin \frac{\pi }{4}=\sin (\pi +\frac{\pi }{4})=\sin \frac{5\pi }{4}$

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

$\sin \theta =-\frac{1}{\sqrt{2}}$ হলে, 

চতুর্থ চতুর্ভাগে, $\sin \theta =-\frac{1}{\sqrt{2}}=-\sin \frac{\pi }{4}=\sin (2\pi -\frac{\pi }{4})=\sin \frac{7\pi }{4}$

$\therefore \theta =\frac{7\pi }{4}$

নির্ণেয় মান $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

দেওয়া আছে,

$Q=\frac{-\sin (-\theta )+\cos (-\theta )}{\sec (-\theta )+\tan (-\theta )}\tan \theta =\frac{5}{12}$ এবং $\cos \theta$ ঋণাত্মক হওয়ায় $\theta$ কোণের অবস্থান তৃতীয় চতুর্ভাগে।

অর্থাৎ, $\tan \theta =\frac{5}{12}=\frac{y}{x}$

$\therefore x=12,y=5$

$\therefore r=\sqrt{x^2+y^2}=\sqrt{(12{)}^2+(5{)}^2}$

$=\sqrt{144+25}=\sqrt{169}=13$

$\therefore \sin \theta =\frac{-y}{r}=-\frac{5}{13}$

$\cos \theta =\frac{-x}{r}=-\frac{12}{13}$ এবং $\sec \theta =\frac{1}{\cos \theta }=\frac{1}{-\frac{12}{13}}=-\frac{13}{12}$

এখন, $Q=\frac{-\sin (-\theta )+\cos (-\theta )}{\sec (-\theta )+\tan (-\theta )}$

এবং$=\frac{\sin \theta +\cos \theta }{\sec \theta -\tan \theta }\left[\begin{array}{c}\because \sin (-\theta )=-\sin \theta \\ \cos (-\theta )=\cos \theta \\ \sec (-\theta )=\sec \theta \\ \text{ এবং }\tan (-\theta )=-\tan \theta \end{array}\right]$

$=\frac{-\frac{5}{13}-\frac{12}{13}}{-\frac{13}{12}-\frac{5}{12}}=\frac{\frac{-17}{13}}{\frac{-18}{12}}=-\frac{17}{13}\times (-\frac{12}{18})=\frac{34}{39}$

$\therefore \theta =\frac{34}{39}$. (দেখানো হলো)