সৃজনশীলঃ ‘P=tan⁡A⋅tan⁡B,q=tan⁡C⋅tan⁡Dr=4sin⁡α2sin⁡β2cos⁡γ2−1…’ রাজশাহী বোর্ড ২০১৯ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $p=\tan A\cdot \tan B$ এবং $q=\tan C\cdot \tan DA={20}^{\circ },B=2A,C=3A,D=4A$

বামপক্ষ, $=pq$

$=\tan A\cdot \tan B\cdot \tan C\cdot \tan D$

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

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

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

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

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

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

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

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

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

$\therefore pq=3\cdot$ (দেখানো হলো)

দেওয়া আছে, $\alpha +\beta +\gamma =0$ এবং $r=4\sin \frac{\alpha }{2}\cdot \sin \frac{\beta }{2}\cdot \cos \frac{\gamma }{2}-1$

বামপক্ষ, $=\cos \alpha +\cos \beta -\cos \gamma$

$=2\cos \frac{\alpha +\beta }{2}\cdot \cos \frac{\alpha -\beta }{2}-\cos \gamma$

$=2\cos \left(\frac{-\gamma }{2}\right)\cdot \cos \frac{\alpha -\beta }{2}-\cos \gamma$

$=2\cos \frac{\gamma }{2}\cdot \cos \frac{\alpha -\beta }{2}-\cos \gamma +1-1$

$=2\cos \frac{\gamma }{2}\cdot \cos \frac{\alpha -\beta }{2}-(1+\cos \gamma )+1$

$=2\cos \frac{\gamma }{2}\cdot \cos \frac{\alpha -\beta }{2}-2{\cos }^2\frac{\gamma }{2}+1$

$=2\cos \frac{\gamma }{2}[\cos \frac{\alpha -\beta }{2}-\cos \frac{\gamma }{2}]+1$

$=2\cos \frac{\gamma }{2}\cdot [\cos (\frac{\alpha }{2}-\frac{\beta }{2})-\cos (\frac{\alpha }{2}+\frac{\beta }{2})]+1$

$=2\cos \frac{\gamma }{2}\cdot 2\sin \frac{\alpha }{2}\cdot \sin \frac{\beta }{2}+1$

$=4\sin \frac{\alpha }{2}\cdot \sin \frac{\beta }{2}\cdot \cos \frac{\gamma }{2}+1$

$=4\sin \frac{\alpha }{2}\cdot \sin \frac{\beta }{2}\cdot \cos \frac{\gamma }{2}-1+2$

$=r+2=$ ডানপক্ষ

$\therefore \cos \alpha +\cos \beta -\cos \gamma =r+2$ (প্রমাণিত)