সৃজনশীলঃ ‘…’ ঢাকা বোর্ড ২০২৩ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

উদ্দীপক হতে পাই,

$\angle ABC={40}^{\circ },\angle ACB={120}^{\circ }$

$\therefore \angle A={180}^{\circ }-(\angle ABC+\angle ACB)$

$={180}^{\circ }-({120}^{\circ }+{40}^{\circ })$

$={180}^{\circ }-{160}^{\circ }$

$={20}^{\circ }$

$\therefore$ L.H.S $=\frac{\sqrt{3}}{4}\mathrm{cosec}A-\frac{1}{4}\sec A$

$=\frac{1}{2}(\frac{\sqrt{3}}{2}\mathrm{cosec}{20}^{\circ }-\frac{1}{2}\sec {20}^{\circ })$

$=\frac{1}{2}(\frac{\sin {60}^{\circ }}{\sin {20}^{\circ }}-\frac{\cos {60}^{\circ }}{\cos {20}^{\circ }})$

$=\frac{1}{2}\left(\frac{\sin {60}^{\circ }\cos {20}^{\circ }-\cos {60}^{\circ }\sin {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}\right)$

$=\frac{1}{2}\frac{\sin {40}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}$

$=\frac{1}{2}\frac{2\cdot \sin {20}^{\circ }\cos {20}^{\circ }}{\sin {20}^{\circ }\cos {20}^{\circ }}$

$=1=$ R.H.S (দেখানো হল)

'খ' হতে পাই, $\angle A={20}^{\circ }$

উদ্দীপক হতে পাই, $\theta ={180}^{\circ }-{120}^{\circ }={60}^{\circ }$.

$\therefore \text{ LH.S }=3-{\cos }^2(\theta +A)-{\cos }^{2}A-{\cos }^2(\theta -A)$

$=3-{\cos }^2({60}^{\circ }+{20}^{\circ })-{\cos }^2{20}^{\circ }-{\cos }^2({60}^{\circ }-{20}^{\circ })$

$=3-{\cos }^2{80}^{\circ }-{\cos }^2{20}^{\circ }-{\cos }^2{40}^{\circ }$

$=3-\frac{1}{2}(2{\cos }^2{80}^{\circ }+2{\cos }^2{20}^{\circ }+2{\cos }^2\cdot {40}^{\circ })$

$=3-\frac{1}{2}(1+\cos {160}^{\circ }+1+\cos {40}^{\circ }+1+\cos {80}^{\circ })$

$=3-\frac{1}{2}(3+\cos {160}^{\circ }+2\cos {60}^{\circ }\cos {20}^{\circ })$

$=3-\frac{1}{2}(3+\cos {160}^{\circ }+\cos {20}^{\circ })$

$=3-\frac{1}{2}(3+2\cos {90}^{\circ }\cos {70}^{\circ })$

$=3-\frac{1}{2}\times 3$

$=\frac{3}{2}=R.H.S$ (দেখানো হলো)