সৃজনশীলঃ ‘দৃশ্যকল্প-১ : 4cos3⁡…’ বরিশাল বোর্ড ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\theta ={60}^{\circ }$

এবং $4{\cos }^{3}A-\cos \theta =3\cos A$

বা, $4{\cos }^{3}A-3\cos A=\cos \theta$

বা, $\cos 3A=\cos {60}^{\circ }$

বা, $3A={60}^{\circ }$

$\therefore A={20}^{\circ }$

বামপক্ষ $=\tan A\tan 2A\tan 3A\tan 4A$

$=\tan {20}^{\circ }\tan {40}^{\circ }\tan {60}^{\circ }\tan {80}^{\circ }$

$=\sqrt{3}\cdot \tan {20}^{\circ }\cdot \tan {40}^{\circ }\tan {80}^{\circ }$

$=\sqrt{3}\cdot \tan {20}^{\circ }\cdot \tan ({60}^{\circ }-{20}^{\circ })\cdot \tan ({60}^{\circ }+{20}^{\circ })$

$=\sqrt{3}\cdot \tan {20}^{\circ }\cdot \frac{\tan {60}^{\circ }-\tan {20}^{\circ }}{1+\tan {60}^{\circ }\tan {20}^{\circ }}\cdot \frac{\tan {60}^{\circ }+\tan {20}^{\circ }}{1-\tan {60}^{\circ }\tan {20}^{\circ }}$

$=\sqrt{3}\cdot \tan {20}^{\circ }\cdot \left(\frac{\sqrt{3}-\tan {20}^{\circ }}{1+\sqrt{3}\tan {20}^{\circ }}\right)\left(\frac{\sqrt{3}+\tan {20}^{\circ }}{1-\sqrt{3}\tan {20}^{\circ }}\right)$

$=\sqrt{3}\cdot \tan {20}^{\circ }\cdot \frac{3-{\tan }^2{20}^{\circ }}{1-3{\tan }^2{20}^{\circ }}$

$=\sqrt{3}\cdot \frac{3\tan {20}^{\circ }-{\tan }^3{20}^{\circ }}{1-3{\tan }^2{20}^{\circ }}$

$=\sqrt{3}\cdot \tan \left({3.20}^{\circ }\right)$

$=\sqrt{3}\cdot \tan {60}^{\circ }=\sqrt{3}\cdot \sqrt{3}$

= 3 = ডানপক্ষ

$A\tan 2A\tan 3A\tan 4A=3.$ (দেখানো হলো)

দেওয়া আছে, $\cot A+\cot B+\cot C=\cot A\cot B\cot C$

বা, $\cot A+\cot B=\cot A\cot B\cot C-\cot C$

বা, $\cot A+\cot B=\cot C(\cot A\cot B-1)$

বা, $\frac{1}{\cot C}=\frac{\cot A\cot B-1}{\cot B+\cot A}$

বা, $\frac{1}{\cot C}=\cot (A+B)$

বা, $\tan C=\cot (A+B)$

বা, $\cot (\frac{\pi }{2}-C)=\cot (A+B)$

বা, $\frac{\pi }{2}-C=A+B$

$\therefore A+B+C=\frac{\pi }{2}$

বামপক্ষ $={\cos }^{2}A+{\cos }^{2}B+{\sin }^{2}C$

$=\frac{1}{2}[2{\cos }^{2}A+2{\cos }^{2}B]+{\sin }^{2}C$

$=\frac{1}{2}(1+\cos 2A+1+\cos 2B)+{\sin }^{2}C$

$=\frac{1}{2}\cdot 2+\frac{1}{2}(\cos 2A+\cos 2B)+{\sin }^{2}C$

$=1+\frac{1}{2}\cdot 2\cos (A+B)\cos (A-B)+{\sin }^{2}C$

$=1+\cos (\frac{\pi }{2}-C)\cos (A-B)+{\sin }^{2}C$

$=1+\sin C\cos (A-B)+{\sin }^{2}C$

$=1+\sin C[\cos (A-B)+\sin C]$

$=1+\sin C[\cos (A-B)+\sin \{\frac{\pi }{2}-(A+B\left)\right\}]$

$=1+\sin C[\cos (A-B)+\cos (A+B\left)\right]$

$=1+\sin C.2\cos A\cos B$

$=1+2\cos A\cos B\sin C$

= ডানপক্ষ

$\therefore {\cos }^{2}A+{\cos }^{2}B+{\sin }^{2}C$$=1+2\cos A\cos B\sin C$ (প্রমাণিত)