সৃজনশীলঃ ‘দৃশ্যকল্প-১ :দৃশ্যকল্প-২ :…’ সিলেট এমসি কলেজ ২০২৪ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দৃশ্যকল্প-১ এ

$B{C}^2=A{C}^2-A{B}^2$

$\Rightarrow B{C}^2=(2\sqrt{3}{)}^2-(\sqrt{6}-1{)}^2$

$\Rightarrow B{C}^2=12-(6-2\sqrt{6}+1)$

$\Rightarrow B{C}^2=12-7+2\sqrt{6}$

$\Rightarrow B{C}^2=5+2\sqrt{6}$

$\Rightarrow BC=\sqrt{5+2\sqrt{6}}$

$=\sqrt{3+2\sqrt{3}\cdot \sqrt{2}+2}$

$=\sqrt{(\sqrt{3}{)}^2+2\cdot \sqrt{3}\cdot \sqrt{2}+(\sqrt{2}{)}^2}$

$=\sqrt{(\sqrt{3}+\sqrt{2}{)}^2}$

$\therefore BC=\sqrt{3}+\sqrt{2}$

$\therefore \tan \alpha =\frac{\sqrt{6}-1}{\sqrt{3}+\sqrt{2}}$

$\therefore \alpha ={\tan }^{-1}\frac{\sqrt{6}-1}{\sqrt{3}+\sqrt{2}}$

এবং,

$AB=\sqrt{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}$

= 1

$\therefore \tan \beta =\frac{1}{\sqrt{2}}$

$\therefore \beta ={\tan }^{-1}\frac{1}{\sqrt{2}}$

$\therefore \text{ L.HS. }=\alpha +\beta$

$={\tan }^{-1}\frac{\sqrt{6}-1}{\sqrt{3}+\sqrt{2}}+{\tan }^{-1}\frac{1}{\sqrt{2}}$

$={\tan }^{-1}\frac{\frac{\sqrt{6}-1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{2}}}{1-\frac{\sqrt{6}-1}{\sqrt{3}+\sqrt{2}}\cdot \frac{1}{\sqrt{2}}}$

$={\tan }^{-1}\frac{\frac{\sqrt{12}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{\sqrt{2}(\sqrt{3}+\sqrt{2})}}{\frac{\sqrt{6}+\sqrt{4}-\sqrt{6}+13}{\sqrt{2}(\sqrt{3}+\sqrt{2})}}$

$={\tan }^{-1}\frac{\frac{2\sqrt{3}+\sqrt{3}}{\sqrt{2}(\sqrt{3}+\sqrt{2})}}{2+1}$

$={\tan }^{-1}\frac{3\sqrt{3}}{\sqrt{2}(\sqrt{3}+\sqrt{2})}$

$={\tan }^{-1}\sqrt{3}$

$={60}^{\circ }$

L.H.S = R.H.S

$\therefore \alpha +\beta ={60}^{\circ }$ (প্রমাণিত)

দৃশ্যকল্প-২ হতে পাই,

$\therefore A{B}^2=A{C}^2-B{C}^2$

$=(5{)}^2-(4{)}^2$

$=25-16$

= 9

$\therefore AB=3$

$\therefore \tan \theta =\frac{3}{4}$

এখন,

${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x=\tan 8$

$\Rightarrow {\sin }^{3}x\cos 3x+{\cos }^{3}x\cdot \sin 3x=\frac{2}{4}$

$\Rightarrow 4{\sin }^{2}x\cdot \cos 3x+4{\cos }^{3}x\cdot \sin 3x=3$

$\Rightarrow (2\sin \theta -\sin 3x)\cos 3x+(\cos 3x+3\cos x)\sin 3x=3$

$\Rightarrow 3\cos 3x\sin x-\sin 3x\cos 3x+\sin 3x\cos 3x+3\sin 3x\cos x=3$

$\Rightarrow 3\cos 3x\cdot \sin x+3\sin 3x\cdot \cos x=3$

$\Rightarrow 3(\sin 3x\cdot \cos x+\cos 3x\cdot \sin x)=3$

$\Rightarrow 3\sin (3x+x)=3$

$\Rightarrow 3\sin 4x=3$

$\Rightarrow \sin 4x=1$

$\Rightarrow 4x=(4n+1)\frac{\pi }{2}$

$\therefore x=(4n+1)\frac{\pi }{8}$; যখন n এর মান শূন্য অথবা যেকোনো পূর্ণ সংখ্যা।