সংক্ষিপ্ত প্রশ্ন ও সমাধান

মনে করি, ABC সমকোণী ত্রিভুজের সবচেয়ে ছোট বাহুর দৈর্ঘ্য BC = 7 সে.মি. এবং সবচেয়ে ছোট কোণ $\angle$BAC = 15°

ত্রিভুজটির অতিভুজ AB = ?

এখন, $△$ABC হতে পাই,

$\sin \angle BAC=\frac{BC}{AB}$

বা, $\sin 15°= \frac{7}{AB}$

বা, $AB = \frac{7}{\sin 15°}$

বা, $AB = 7\sqrt{6}+ 7\sqrt{2}= 7 \left(\sqrt{6}+\sqrt{2}\right)$

$=7\left(\sqrt{3} . \sqrt{2} + \sqrt{2}\right)= 7\sqrt{2} \left(\sqrt{3}+1\right)$

= $\frac{7\sqrt{2} \left(\sqrt{3}+ 1\right) \left(\sqrt{3} - 1\right)}{\sqrt{3} - 1}$

$=7\sqrt{2} \frac{\left\{{\left(\sqrt{3}\right)}^2-1\right\}}{\sqrt{3}-1}=\frac{7\sqrt{2}\times 2}{\sqrt{3}-1}$

$\therefore$ $AB=\frac{14\sqrt{2}}{\sqrt{3}- 1}$

$\therefore$ ত্রিভুজটির অতিভুজের দৈর্ঘ্য $\frac{14\sqrt{2}}{\sqrt{3}- 1}$ সে. মি.।

$\cos \left(-\frac{25\mathrm{\pi }}{3}\right)=\cos \left(\frac{25\mathrm{\pi }}{3}\right)$

$= \cos \left(8\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right)$

$= \cos \left(16 \times \frac{\mathrm{\pi }}{2}+ \frac{\mathrm{\pi }}{3}\right)$

এখানে, n = 16 এবং $\left(16 \times \frac{\mathrm{\pi }}{2} +\frac{\mathrm{\pi }}{3}\right)$ প্রথম চতুর্ভাগে অবস্থিত।

$\cos \left(16\times \frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{3}\right)= \cos \frac{\mathrm{\pi }}{3} = \frac{1}{2}$

নির্ণেয় মান :  $\frac{1}{2}$

$\tan \left(-\frac{25\mathrm{\pi }}{6}\right)$ = $- \tan \left(\frac{25\mathrm{\pi }}{6}\right)$

= $- \tan \left(4\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)$

$= - \tan \left(8 \times \frac{\mathrm{\pi }}{2} + \frac{\mathrm{\pi }}{6}\right)$

এখানে, n = 8 এবং, $\left(8\times \frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{6}\right)$ প্রথম চতুর্ভাগে, অবস্থিত।

= $- \tan \left(8 \times \frac{\mathrm{\pi }}{2} +\frac{\mathrm{\pi }}{6}\right)= - \tan \frac{\mathrm{\pi }}{6}= - \frac{1}{\sqrt{3}}$

নির্ণেয় মান : $- \frac{1}{\sqrt{3}}$

$\tan \frac{7\mathrm{\pi }}{6}=\tan \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right) = \tan \frac{\mathrm{\pi }}{6} = \frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$

$\sin \left(- {750}^{°}\right) = - \sin {750}^{°}$

$= - \sin (8 \times 90°+ 30°)$

$=- \sin 30°=-\frac{1}{2}$

নির্ণেয় মান : $-\frac{1}{2}$

$\tan \left(-\frac{11\mathrm{\pi }}{6}\right)$ = $- \tan \left(\frac{11\mathrm{\pi }}{6}\right)$

$=-\tan \left(\frac{4\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}\right)$

$=-\tan \left(-\tan \frac{\mathrm{\pi }}{6}\right) = \tan \frac{\mathrm{\pi }}{6}=\frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$

$sec\left(-\frac{35\mathrm{\pi }}{6}\right)= sec\frac{25\mathrm{\pi }}{6} \left[sec(- \theta ) = sec \theta \right]$

$= sec \left(8 \times \frac{\mathrm{\pi }}{2} + \frac{\mathrm{\pi }}{6}\right)= sec \frac{\mathrm{\pi }}{6}=\frac{2}{\sqrt{3}}$

এবং $cot\left(-\frac{23\mathrm{\pi }}{3}\right)=-cot \frac{23\mathrm{\pi }}{3} \left[cot(- \theta )=-cot \theta \right]$

$= - cot \left(16 \times \frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{3}\right)= -\left(- cot\frac{\mathrm{\pi }}{3}\right)= cot\frac{\mathrm{\pi }}{3} = \frac{1}{\sqrt{3}}$

নির্ণেয় মান $\frac{2}{\sqrt{3}}$ এবং $\frac{1}{\sqrt{3}}$

দেওয়া আছে, $Q=\frac{{\mathrm{\pi }}^2}{12}=\frac{{\left(180°\right)}^2}{12}$ = 2700°

প্রদত্ত রাশি = tan Q tan 5Q tan 7Q tan 11Q

=$\left(\tan {2700}^{°}\right) \left\{\tan \left(5 \times {2700}^{°}\right)\right\} \left\{\tan \left(7 \times {2700}^{°}\right)\right\} \left\{\tan \left(11 \times {2700}^{°}\right)\right\}$

$=tan(90°\times 30+0°) tan(\left(90° \times 50+0°\right) tan(90°\times 210+0°) tan(90°\times 330+0°)$

$= 0 \times 0 \times 0 \times 0 = 0$

$cot \left(\theta -\frac{9\mathrm{\pi }}{2}\right) = cot \left\{-\left(\frac{9\mathrm{\pi }}{2}-\theta \right)\right\}$

= $- cot \left(9 \times \frac{\mathrm{\pi }}{2}-\theta \right)$

= - tan $\theta$

নির্ণেয় মান : = - tan $\theta$

প্রদত্ত রাশি $= \tan \frac{7\mathrm{\pi }}{6}= \tan \left(2 \times \frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{6} \right) = \tan \frac{\mathrm{\pi }}{6} =\frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$

দেওয়া আছে, $\sin \theta = - \frac{\sqrt{3}}{2}; \frac{\mathrm{\pi }}{2}<0<\frac{3\mathrm{\pi }}{2}$

তৃতীয় চতুর্ভাগে $\sin \theta =-\frac{\sqrt{3}}{2}$

বা, $\sin \theta = - \sin \frac{\mathrm{\pi }}{3}$

বা, $\sin \theta = \sin \left(\mathrm{\pi }+ \frac{\mathrm{\pi }}{3}\right)= \sin \left(\frac{3\mathrm{\pi }+\mathrm{\pi }}{3}\right)$

$\theta = \frac{4\mathrm{\pi }}{3} ;$ যা $\frac{\mathrm{\pi }}{2}< \theta <\frac{3\mathrm{\pi }}{2}$ শর্ত পূরণ করে।

নির্ণেয় মান : $\theta = \frac{4\mathrm{\pi }}{3}$

দেওয়া আছে, $\sin A = - \frac{1}{\sqrt{2}}$ ; যেখানে  $\theta <A<\frac{3\mathrm{\pi }}{2}$

তৃতীয় চতুর্ভাগে, $\sin A =-\frac{1}{\sqrt{2}}$

বা, $\sin A = - \sin \frac{\mathrm{\pi }}{4}= \sin \left(\pi + \frac{\mathrm{\pi }}{4}\right) = \sin \frac{5\mathrm{\pi }}{4}$

$\therefore$ $A=\frac{5\mathrm{\pi }}{4}$

নির্ণেয় মান : $\frac{5\mathrm{\pi }}{4}$

দেওয়া আছে, $\cos \theta =-\frac{1}{2}$ যেখানে, $\mathrm{\pi }<\theta <\frac{3\mathrm{\pi }}{2}$

$\mathrm{\pi }<\theta <\frac{3\mathrm{\pi }}{2}$ ব্যবধিতে $\theta$ তৃতীয় চতুর্ভাগে অবস্থান করে। তৃতীয় চতুর্ভাগে cos $\theta$ ঋণাত্মক।

$\cos \theta =-\frac{1}{2}= - \cos \frac{\mathrm{\pi }}{3}= \cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{3}\right) = \cos \frac{4\mathrm{\pi }}{3}$

$\therefore$ $\theta =\frac{4\mathrm{\pi }}{3}$

নির্ণেয় মান : $\theta =\frac{4\mathrm{\pi }}{3}$

দেওয়া আছে, tan 10x = cot 5x

বা, $\tan (2 \times 5x) = \frac{1}{\tan 5x}$

বা, $l\frac{2 \tan 5x}{1-{(\tan 5x)}^2}= \frac{1}{\tan 5x}$

বা, $2ta{n}^{2}5x=1-ta{n}^{2}5x$

বা, $2ta{n}^{2}5x+ta{n}^{2}5x=1$

বা, $3ta{n}^{2}5x=1$

বা, $ta{n}^{2}5x=\frac{1}{3}$

বা, $\tan 5x =\pm \frac{1}{\sqrt{3}}$ [বর্গমূল করে]

হয়, $\tan 5x=\frac{1}{\sqrt{3}}= \tan \frac{\mathrm{\pi }}{6}, \tan \left(\pi +\frac{\mathrm{\pi }}{6}\right)$

বা, $5x = \frac{\mathrm{\pi }}{6}, \frac{7\mathrm{\pi }}{6}$

$\therefore$ $x =\frac{\mathrm{\pi }}{30},\frac{7\mathrm{\pi }}{30}$

অথবা, $\tan 5x =-\frac{1}{\sqrt{3}}$

বা, $\tan 5x = - \tan \frac{\mathrm{\pi }}{6}$

বা, $\tan 5x = \tan \left(\mathrm{\pi }- \frac{\mathrm{\pi }}{6}\right), \left(2\mathrm{\pi }- \frac{\mathrm{\pi }}{6}\right)$

বা, $5x=\frac{5\mathrm{\pi }}{6}, \frac{11\mathrm{\pi }}{6}$

$\therefore$ $x=\frac{\mathrm{\pi }}{6}$ বা $\frac{11\mathrm{\pi }}{30}$

নির্ণেয় মান : $x =\frac{\mathrm{\pi }}{30},\frac{7\mathrm{\pi }}{30}$, $\frac{\mathrm{\pi }}{6}$, $\frac{11\mathrm{\pi }}{30}$

দেওয়া আছে, $\sin \left(\theta -\frac{3\mathrm{\pi }}{2}\right)=\frac{\sqrt{3}}{2}$

বা, $sin\left\{-\left(\frac{3}{2}-\theta \right)\right\}=\frac{\sqrt{3}}{2}$

বা, $- \sin \left(\frac{3\mathrm{\pi }}{2}-\theta \right) = \frac{\sqrt{3}}{2}$

বা, $- (- \cos \theta ) =\frac{\sqrt{3}}{2}$

বা, $\cos \theta =\frac{\sqrt{3}}{2}= \cos \frac{\mathrm{\pi }}{6}$

বা, $\cos \left(2\mathrm{\pi }- \frac{\mathrm{\pi }}{6}\right)$

$\theta =\frac{\mathrm{\pi }}{6}$ বা $\frac{11\mathrm{\pi }}{6}$

$\theta =\frac{\mathrm{\pi }}{6}$ হলে, $\tan \theta = \tan \frac{\mathrm{\pi }}{6}=\frac{1}{\sqrt{3}}$

$\theta =\frac{11\mathrm{\pi }}{6}$ হলে, $\tan \theta = \tan \frac{11\mathrm{\pi }}{6}= - \frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$ অথবা - $\frac{1}{\sqrt{3}}$

দেওয়া আছে, $\cos ec A =-\frac{5}{3}$

বা, $\cos e{c}^2 A ={\left(-\frac{5}{3}\right)}^2$ [বর্গ করে]

বা,  $\cos e{c}^2 A =\frac{25}{9}$

বা, $\cos e{c}^2 A -1 =\frac{25}{9}-1$

বা, $co{t}^{2}A =\frac{25-9}{9}=\frac{16}{9}$

বা, $cot A=\pm \sqrt{\frac{16}{9}}=\pm \frac{4}{3}=\frac{4}{3}$

নির্ণেয় মান : $\frac{4}{3}$

দেওয়া আছে, $\sin A=\frac{\sqrt{3} }{2}$ যেখানে $\frac{\mathrm{\pi }}{2}\le A<\frac{3\mathrm{\pi }}{2}$

বা, $\sin A = \sin \frac{\mathrm{\pi }}{3}= \sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)=\sin \frac{\mathrm{\pi }}{3}= \sin \frac{2\mathrm{\pi }}{3}$

$\therefore$ $A=\frac{2\mathrm{\pi }}{3}$

নির্ণেয় মান : $\frac{2\mathrm{\pi }}{3}$

বামপক্ষ = $\cos \left(\frac{17\mathrm{\pi }}{10}\right)+ \cos \left(\frac{13\mathrm{\pi }}{10}\right)+ \cos \left(\frac{9\mathrm{\pi }}{10}\right)+ \cos \left(\frac{\mathrm{\pi }}{10}\right)$

= $\cos \left(\frac{17\mathrm{\pi }}{10}\right)+ \cos \left(3\mathrm{\pi }-\frac{17\mathrm{\pi }}{10}\right)+ \cos \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{10}\right)+ \cos \left(\frac{\mathrm{\pi }}{10}\right)$

= $\cos \frac{17\mathrm{\pi }}{10}- \cos \frac{17\mathrm{\pi }}{10}- \cos \frac{\mathrm{\pi }}{10}+ \cos \frac{\mathrm{\pi }}{10}$

= 0 = ডানপক্ষ

$\therefore$ $\cos \left(\frac{17\mathrm{\pi }}{10}\right)+ \cos \left(\frac{13\mathrm{\pi }}{10}\right)+ \cos \left(\frac{9\mathrm{\pi }}{10}\right)+ \cos \left(\frac{\mathrm{\pi }}{10}\right)$ = 0

$\tan \left(-\frac{11\mathrm{\pi }}{6}\right)=-\tan \frac{11\mathrm{\pi }}{6}$

$=-\tan \left(\frac{4\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}\right)$

$= -\left( - \tan \frac{\mathrm{\pi }}{6}\right)$

$= \tan \frac{\mathrm{\pi }}{6}= \frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$

$cot \left(-480°\right)=-cot 480°$

$= - cot (5 \times 90°+ 30°)$

$= - (- \tan 30°)$

$= \tan 30°=\frac{1}{\sqrt{3}}$

নির্ণেয় মান : $\frac{1}{\sqrt{3}}$

$\frac{\mathrm{\pi }}{2}<\theta <\mathrm{\pi }$ ব্যবধিতে $\theta$এর অবস্থান দ্বিতীয় চতুর্ভাগে।

$tan \theta =-\sqrt{3}$ $= - \tan 60°= \tan (180° - 60°)$

$\therefore$ $\theta = 120°=\frac{2\mathrm{\pi }}{3}$

প্রদত্ত রাশি $= sec(- 225°)$

$= sec 225°$ $\left[sec(-\theta ) = sec\left(\theta \right)\right]$

$= sec(2 \times 90°+ 45°)$

= - sec 45°

$=-\sqrt{2}$

প্রদত্ত রাশি $= \cos \left(- \frac{35\mathrm{\pi }}{6}\right)= \cos \left(\frac{35\mathrm{\pi }}{6}\right)$

$= \cos \left(11 \times \frac{\mathrm{\pi }}{2}+ \frac{\mathrm{\pi }}{3}\right)$

$=\sin \frac{\mathrm{\pi }}{3 }=\frac{\sqrt{3}}{2}$

এখানে,

$\tan \left(\frac{\mathrm{n\pi }}{2}+\theta \right)= - \frac{1}{\sqrt{3}}$

বা, $\tan \frac{n\mathrm{\pi }}{2}+ \theta = - \tan \frac{\mathrm{\pi }}{6}$

বা, $t\mathrm{an}\frac{n\mathrm{\pi }}{2}+ \theta = \tan \left(\mathrm{\pi } -\frac{\mathrm{\pi }}{6}\right)$

বা, $t\mathrm{an}\left(\frac{n\mathrm{\pi }}{2}+ \theta \right) + \tan \frac{5\mathrm{\pi }}{6}$

বা, $\frac{\mathrm{n\pi }}{2}+ \frac{\mathrm{\pi }}{3} = \frac{5\mathrm{\pi }}{6}$

বা, $\frac{\mathrm{n\pi }}{2}= \frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{3}$

বা, $\frac{\mathrm{n\pi }}{2}= \frac{5\mathrm{\pi } - 2\mathrm{\pi }}{6}$

বা, $\frac{\mathrm{n\pi }}{2}= \frac{3\mathrm{\pi }}{6}$

বা, $n= \frac{3\mathrm{\pi }\times 2}{6\times \mathrm{\pi }}$

$\therefore$ n = 1

দেওয়া আছে,

$\cos (\mathrm{\pi }+ x) = -\frac{\sqrt{3}}{2}$

বা, $\cos \left(2 \times \frac{\mathrm{\pi }}{2}+ x\right)= - \frac{\sqrt{3}}{2}$

বা, $- \cos x = -\frac{\sqrt{3}}{2}$

বা, $\cos x = \frac{\sqrt{3}}{2}$

বা, $\cos x = \cos \frac{\mathrm{\pi }}{6}$

$\therefore$ $x = \frac{\mathrm{\pi }}{6}$

প্রদত্ত রাশি $= \tan \left(\frac{3\mathrm{\pi }}{2}+\theta \right) = \tan \left(3 \times \frac{\mathrm{\pi }}{2}+\theta \right)$

tan $\theta$ পড়বে চতুর্থ চতুর্ভাগে যেখানে tan $\theta$ ঋণাত্মক। $\frac{\mathrm{\pi }}{2}$ এর সাথে বিজোড় সংখ্যা 3 দ্বারা গুণ আছে বিধায় tan $\theta$ এর জায়গায় – cot $\theta$ হবে।

$\therefore$ $\tan \left(\frac{3\mathrm{\pi }}{2}+\theta \right) = - cot \theta$

এখানে, $cot \left(\frac{\mathrm{n\pi }}{4}+ \theta \right)= 1 .$

বা, $cot \left(\frac{\mathrm{n\pi }}{4}- \frac{\mathrm{\pi }}{2}\right)= cot \frac{\pi }{4}$

বা, $\frac{n\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{2}= \frac{\mathrm{\pi }}{4}$

বা, $\frac{n\mathrm{\pi }}{4}=\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{2}=\frac{3\mathrm{\pi }}{4}$

$\therefore$ n = 3