সৃজনশীলঃ ‘দৃশ্যকল্প-১ : θ=6∘দৃশ্যকল্প-২…’ বিয়োজন যোজন করে ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\theta =6^{\circ }$

বামপক্ষ $=\tan \theta \cdot \tan 7\theta \tan 11\theta \tan 13\theta$

$=\tan 6^{\circ }\tan (7\times 6^{\circ })\tan (11\times 6^{\circ })\tan (13\times 6^{\circ })$

$=\tan 6^{\circ }\tan {42}^{\circ }\tan {66}^{\circ }\tan {78}^{\circ }$

$=\frac{\sin 6^{\circ }\sin {42}^{\circ }\sin {66}^{\circ }\sin {78}^{\circ }}{\cos 6^{\circ }\cos {42}^{\circ }\cos {66}^{\circ }\cos {78}^{\circ }}$

$=\frac{2\sin {66}^{\circ }\sin 6^{\circ }-2\sin {78}^{\circ }\sin {42}^{\circ }}{2\cos {66}^{\circ }\cos 6^{\circ }\cdot 2\cos {78}^{\circ }\cos {42}^{\circ }}$

$=\frac{\cos {60}^{\circ }-\cos {72}^{\circ }}{\cos {72}^{\circ }+\cos {60}^{\circ }}\cdot \frac{\cos {36}^{\circ }-\cos {120}^{\circ }}{\cos {120}^{\circ }+\cos {36}^{\circ }}$

$=\frac{\frac{1}{2}-\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}+1}{4}+\frac{1}{2}}\cdot \frac{\frac{\sqrt{5}-1}{4}+\frac{1}{2}}{-\frac{1}{2}+\frac{\sqrt{5}-1}{4}}$

$=\frac{2-\sqrt{5}-1}{\sqrt{5}+1+2}\cdot \frac{\sqrt{5}-1+2}{-2+\sqrt{5}-1}$

$=\frac{(1-\sqrt{5})(1+\sqrt{5})}{(3+\sqrt{5})(\sqrt{5}-3)}=\frac{1-5}{5-9}=\frac{-4}{-4}=1=$ ডाনপক্ষ

$\therefore \tan \theta \cdot \tan 7\theta \tan 11\theta \tan 13\theta =1$. (প্রমাণিত)

দেওয়া আছে,

$f\left(x\right)=\cos x,g\left(x\right)=\sin x$ এবং $h\left(x\right)=\tan x$

এখানে, $h(\frac{\pi }{2}-A)h(\frac{\pi }{2}-B)h(\frac{\pi }{2}-C)=h(\frac{\pi }{2}-D)$

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

বা, $\cot A\cdot \cot B\cdot \cot C=\cot D$

অর্থাৎ দেখাতে হবে যে, $\cot A\cdot \cot B\cdot \cot C=\cot D$.

এখন, $f(A+B)\cdot g(C+D)=f(A-B)g(C-D)$ হলে,

$\cos (A+B)\sin (C+D)=\cos (A-B)\sin (C-D)$

বा, $\frac{\sin (C+D)}{\sin (C-D)}=\frac{\cos (A-B)}{\cos (A+B)}$

বা, $\frac{\sin (C+D)-\sin (C-D)}{\sin (C+D)+\sin (C-D)}=\frac{\cos (A-B)-\cos (A^{\circ }+B)}{\cos (A-B)+\cos (A+B)}$

[বিয়োজন-যোজন করে]

বा, $\frac{2\cos C\sin D}{2\sin C\cos D}=\frac{2\sin A\sin B}{2\cos A\cos B}$

বা, $\frac{\cot C}{\cot D}=\frac{1}{\cot A\cdot \cot B}$

$\therefore \cot A\cdot \cot B\cdot \cot C=\cot D$. (দেখানো হলো)