সৃজনশীলঃ ‘acos2⁡x+bsin2x=c;a>c>b>0 এবং sin⁡θ=513.…’ ময়মনসিংহ বোর্ড ২০২৩ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $a{\cos }^{2}x+b{\sin }^{2}x=c$

বা, $a(1-{\sin }^{2}x)+b{\sin }^{2}x=c$

বা, $a-a{\sin }^{2}x+b{\sin }^{2}x=c$

বা, $(b-a){\sin }^{2}x=c-a$

বা, ${\sin }^{2}x=\frac{c-a}{b-a}$ ....... (i)

আবার,

$a{\cos }^{2}x+b{\sin }^{2}x=c$

বা, $a{\cos }^{2}x+b(1-{\cos }^{2}x)=c$

বা, $a{\cos }^{2}x+b-b{\cos }^{2}x=c$

বা, $(a-b){\cos }^{2}x=c-b$

বা, ${\cos }^{2}x=\frac{c-b}{a-b}$

বা, ${\cos }^{2}x=\frac{b-c}{b-a}$ ....... (ii)

(i) ÷ (ii) করে পাই,

$\frac{{\sin }^{2}x}{{\cos }^{2}x}=\frac{c-a}{b-a}\times \frac{b-a}{b-c}$

বা, ${\tan }^{2}x=\frac{c-a}{b-a}$

$\therefore \tan x=\pm \sqrt{\frac{c-a}{b-c}}$. (প্রমাণিত)

দেওয়া আছে, $\sin \theta =\frac{5}{13}$

বা, ${\sin }^2\theta ={\left(\frac{5}{13}\right)}^2$

বা, $1-{\cos }^2\theta =\frac{25}{169}$

বা, ${\cos }^2\theta =1-\frac{25}{169}$

বা, ${\cos }^2\theta =\frac{144}{169}$

বা, $\cos \theta =\pm \sqrt{\frac{144}{169}}$

বা, $\cos \theta =\pm \frac{12}{13}$

$\therefore \cos \theta =-\frac{12}{13}[\because \cos \theta$ ঋণাত্মক]

বামপক্ষ $=\frac{\tan \theta +\sec (-\theta )}{\cot \theta +\mathrm{cosec}(-\theta )}$

$=\frac{\tan \theta +\sec \theta }{\cot \theta -\mathrm{cosec}\theta }$

$=\frac{\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}{\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }}$

$=\frac{\frac{\frac{5}{13}}{\frac{-12}{13}}+\frac{1}{-\frac{12}{13}}}{\frac{\frac{-12}{13}}{\frac{5}{13}}-\frac{1}{\frac{5}{13}}}$

$=\frac{-\frac{5}{12}-\frac{13}{12}}{-\frac{12}{5}-\frac{13}{5}}$

$=\frac{\frac{-18}{12}}{\frac{-25}{5}}=\frac{-18}{12}\times \frac{5}{-25}$

$=\frac{3}{10}=$ ডানপক্ষ

∴ বামপক্ষ $=$ ডানপক্ষ (দেখানো হলো)