সৃজনশীলঃ ‘এ△ABC−এA=12×0, B=x0,C=5×0.…’ ক্যান্টনমেন্ট পাবলিক স্কুল অ্যান্ড কলেজ ২০২৪ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$△ABC$, এ,

$A=12x^{\circ },B=x^{\circ },C=5x^{\circ }$

$\therefore 12x^{\circ }+x^{\circ }+5x^{\circ }={180}^{\circ }$

$\Rightarrow 18x^{\circ }={180}^{\circ }$

$\therefore x={10}^{\circ }$

$\therefore 12\cdot {10}^{\circ }={120}^{\circ }$

$B={10}^{\circ }$

$C=5\cdot {10}^{\circ }={50}^{\circ }$

L.H.S.

$=\mathrm{cosec}B-\mathrm{se}\sqrt{3}\sec B$

$=\frac{1}{\sin B}(-\frac{\sqrt{3}}{\cos B}$

$=\frac{\cos B-\sqrt{3}\sin B}{\sin B\cdot \cos B}\cdot$

$=\frac{2\cdot \frac{1}{2}\cdot \cos B-\frac{\sqrt{3}}{2}\sin B)}{\frac{1}{2}\cdot 2\sin B\cdot \cos B}$

$=\frac{4\cdot (\sin {30}^{\circ }\cdot \cos B-\cos {30}^{\circ }\cdot \sin B)}{\sin 2B}$

$=\frac{4\cdot \sin ({30}^{\circ }-B)}{\sin 2B}=\frac{4\cdot \sin ({30}^{\circ }-{10}^{\circ })}{\sin 2\cdot {10}^{\circ }}$

$=\frac{4\cdot \sin {20}^{\circ }}{\sin {20}^{\circ }}$

= 4

= R.H.S.

$\therefore \mathrm{cosec}B-5\sqrt{3}\sec B=y$ (showed)

দেওয়া আছে,

$B=x={10}^{\circ }$

‘খ’ হতে, $A={120}^{\circ }$

L.H.S.

$={\sin }^{2}B+{\sin }^2(A+B)+{\sin }^3(2A+B)$

$={\sin }^{2}x+{\sin }^2({120}^{\circ }+x)+{\sin }^2(2\cdot {120}^{\circ }+x)$

$=4{\sin }^{2}x+4{\sin }^2({120}^{\circ }+x)+4{\sin }^2({240}^{\circ }+x))\cdot \frac{1}{4}$

$\begin{array}{c}=\frac{1}{4}\{3\sin x-\sin 3x+3\sin ({120}^{\circ }+x)-\sin 3({120}^{\circ }+x)+3\sin ({240}^{\circ }+x)+3\sin 3({240}^{\circ }+x)\}\end{array}$

= $\frac{1}{4}[3\sin x-\sin 3x+3\{\sin ({120}^{\circ }+x)+\sin ({240}^{\circ }+x)\}-\sin ({360}^{\circ }+3x)-\sin ({720}^{\circ }+3x)]$

= $\frac{1}{4}\{3\sin x-\sin 3x+3\cdot 2\sin ({180}^{\circ }+x)\cos {60}^{\circ }-\sin 3x-\sin 3x\}$

= $\frac{1}{4}(3\sin x-3\sin 3x-3\cdot 2\cdot \sin x\cdot \frac{1}{2})$

= $\frac{1}{4}(3\sin x-3\sin 3x-3\sin x)$

= $-\frac{3}{4}\sin 3x$

= $-\frac{3}{4}\sin 3\cdot {10}^{\circ }$

= $-\frac{3}{4}\sin {30}^{\circ }=\frac{-3}{4}.12$

$=\frac{-3}{8}$ = R.H.S

$\therefore {\sin }^{3}B+{\sin }^3(A+B)+{\sin }^2(2A+B)=\frac{-3}{8}$ (proved).