সৃজনশীলঃ ‘(i) T=sec⁡x+tan⁡x.(ii) M=cos3⁡x+cos3⁡(60∘−x)+cos3⁡(60∘+x)…’ সিলেট বোর্ড ২০২২ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $T=\sec x+\tan x$

$=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

$=\frac{1+\sin x}{\cos x}$

$=\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}$

$=\frac{{(\cos \frac{x}{2}+\sin \frac{x}{2})}^2}{(\cos \frac{x}{2}+\sin \frac{x}{2})(\cos \frac{x}{2}-\sin \frac{x}{2})}$

$=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$

$=\frac{\cos \frac{x}{2}(1+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})}{\cos \frac{x}{2}(1-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})}$

$=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$

$=\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}\cdot \tan \frac{x}{2}t}$

$\Rightarrow T=\tan (\frac{\pi }{4}+\frac{x}{2})$ (proved)

দেওয়া আছে, $M={\cos }^{3}x+{\cos }^3({60}^{\circ }-x)+{\cos }^3({60}^{\circ }+x)$

L.H.S = 4M

$=4({\cos }^{3}x+{\cos }^3({60}^{\circ }-x)+{\cos }^3({60}^{\circ }+x)$

$=4{\cos }^{3}x+4{\cos }^3({60}^{\circ }-x)+4{\cos }^3({60}^{\circ }+x)$

$=3\cos x+\cos 3x+3\cos ({60}^{\circ }-x)+\cos ({180}^{\circ }-3x)+3\cos ({60}^{\circ }+x)+\cos ({180}^{\circ }+3x)$

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

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

$=3\cos x+3\cdot 2\cdot \frac{1}{2}\cdot \cos x-\cos 3x$

$=6\cos x-\cos 3x$

= R. H. S (showed)