সৃজনশীলঃ ‘(i) 5sin2⁡θ+4cos2⁡θ=P(ii) cosecα+cot⁡α=M.…’ দিনাজপুর বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $\mathrm{cosec}\alpha +\cot \alpha =M$

এখন, $M=\sqrt{2}+\sqrt{3}$ হলে,

$\mathrm{cosec}\alpha +\cot \alpha =\sqrt{2}+\sqrt{3}$

বা. $\frac{1}{\sin \alpha }+\frac{\cos \alpha }{\sin \alpha }=\sqrt{2}+\sqrt{3}[\because \mathrm{cosec}\alpha =\frac{1}{\sin \alpha }$ এবং $\cot \alpha =\frac{\cos \alpha }{\sin \alpha }]$

বা, $\frac{1+\cos \alpha }{\sin \alpha }=\sqrt{2}+\sqrt{3}$

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

বা, $\frac{(1+\cos \alpha {)}^2}{{\sin }^2\alpha }=(\sqrt{2}{)}^2+2\cdot \sqrt{2}\cdot \sqrt{3}+(\sqrt{3}{)}^2$

বা, $\frac{(1+\cos \alpha {)}^2}{1-{\cos }^2\alpha }=2+2\sqrt{6}+3[\because {\sin }^2\alpha =1-{\cos }^2\alpha ]$

বা, $\frac{(1+\cos \alpha )(1+\cos \alpha )}{(1+\cos \alpha )(1-\cos \alpha )}=5+2\sqrt{6}$

বা, $\frac{1+\cos \alpha }{1-\cos \alpha }=5+2\sqrt{6}$

বা, $\frac{1+\cos \alpha +1-\cos \alpha }{1+\cos \alpha -1+\cos \alpha }=\frac{5+2\sqrt{6}+1}{5+2\sqrt{6}-1}$ [যোজন-বিয়োজন করে]

বা, $\frac{2}{2\cos \alpha }=\frac{6+2\sqrt{6}}{4+2\sqrt{6}}$

বা, $\frac{1}{\cos \alpha }=\frac{2\sqrt{3}(\sqrt{3}+\sqrt{2})}{2\sqrt{2}(\sqrt{2}+\sqrt{3})}$

বা, $\frac{1}{\cos \alpha }=\frac{\sqrt{3}}{\sqrt{2}}$

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

বা, $\frac{1}{{\cos }^2\alpha }=\frac{3}{2}$

বা, ${\sec }^2\alpha =\frac{3}{2}$

$\therefore \sec \alpha =\pm \sqrt{\frac{3}{2}}$. (প্রমাণিত)

দেওয়া আছে, $5{\sin }^2\theta +4{\cos }^2\theta =P$

এখন, 2P=9 হলে,

$2(5{\sin }^2\theta +4{\cos }^2\theta )=9$

বা, $10{\sin }^2\theta +8{\cos }^2\theta =9$

বা, $10{\sin }^2\theta +8(1-{\sin }^2\theta )=9[\because {\cos }^2\theta =1-{\sin }^2\theta ]$

বা, $10{\sin }^2\theta +8-8{\sin }^2\theta =9$

বা, $2{\sin }^2\theta =9-8$

বা, $2{\sin }^2\theta =1$

বা, ${\sin }^2\theta =\frac{1}{2}$

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

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

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

$0<\theta <2\pi$ ব্যবধিতে $\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

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