\(\sin \theta = \frac{4}{5}\)
\(\therefore \cos \theta = \sqrt {1 - {{\sin }^2}\theta }\)
= \(\sqrt {1 - {{\left( {\frac{4}{5}} \right)}^2}}\)
\(= \sqrt {1 - \frac{{16}}{{25}}}\)
\(= \sqrt {\frac{9}{{25}}}\)
\(=\sqrt{\frac{25-16}{25}}\)
\(= \frac{3}{5}\)
\(\therefore\) \(\frac{{cosec \theta }}{{1 + \cot \theta }}\)
\(=\frac{\frac{1}{\sin \theta}}{1 + \frac{{\cos \theta }}{{\sin \theta }}}\)
\(=\frac{\frac{1}{4 / 5}}{1+\frac{3 / 5}{4 / 5}}=\frac{\frac{5}{4}}{1+\frac{3}{4}}=\frac{\frac{5}{4}}{\frac{4+3}{4}}=\frac{\frac{5}{4}}{\frac{7}{4}}=\frac{5}{4} \times \frac{4}{7}=\frac{5}{7}\)
\(\therefore\) নির্ণেয় মান \(=\frac{5}{7}\)

দেওয়া আছে, \(\tan \theta=\frac{3}{4}\)
বা, \(\cot \theta=\frac{4}{3}\)
বা, \(\cot ^{2} \theta=\frac{16}{9}\) [বর্গ করে]
বা, \(\operatorname{cosec}^{2} \theta-1=\frac{16}{9}\)
বা, \(\operatorname{cosec}^{2} \theta=\frac{16}{9}+1\)
বা, \(\operatorname{cosec}^{2} \theta=\frac{16+9}{9}\)
বা, \(\operatorname{cosec}^{2} \theta=\frac{25}{9}\)
বা, \(\frac{1}{\sin ^{2} \theta}=\frac{25}{9}\)
বা, \(\sin ^{2} 0=\frac{9}{25}\)
বা, \(\sin \theta=\sqrt{\frac{9}{25}}\)
বা, \(\sin \theta=\frac{3}{5}\)
এখন, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}}\)
\(=\sqrt{\frac{\frac{5-3}{5}}{\frac{5+3}{5}}}\)
\(=\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\sqrt{\frac{2}{8}}\)
\(=\sqrt{\frac{1}{4}}=\frac{1}{2}.\)
\(\therefore \sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{1}{2}.\) (প্রমাণিত)

\(\frac{8 \sin \theta+5 \cos \theta}{\sin ^{3} \theta-2 \cos ^{3} \theta+7 \cos \theta}\)
\(=\frac{\left(\frac{8 \sin \theta}{\cos \theta}+\frac{5 \cos \theta}{\cos \theta}\right) \cos \theta}{\left(\frac{\sin ^{3} \theta}{\cos ^{3} \theta}-\frac{2 \cos ^{3} \theta}{\cos ^{3} \theta}+\frac{7 \cos \theta}{\cos ^{3} \theta}\right) \cos ^{3} \theta}\)
\(=\frac{8 \tan \theta+5}{\left(\tan ^{3} \theta-2+\frac{7}{\cos ^{2} \theta}\right) \cos ^{2} \theta}=\frac{(8 \tan \theta+5) \sec ^{2} \theta}{\tan ^{3} \theta-2+7 \sec ^{2} \theta}\)
\(=\frac{(8 \tan \theta+5)\left(1+\tan ^{2} \theta\right)}{\tan ^{3} \theta-2+7\left(1+\tan ^{2} \theta\right)}=\frac{(8 \times 1+5)(1+1)}{1^{3}-2+7\left(1+1^{2}\right)}\left[\because \tan ^{2} \theta=1^{2}=1\right]\)
\(=\frac{(8+5)(2)}{1-2+7(1+1)}=\frac{13 \times 2}{1-2+7 \times 2}\)
\(=\frac{26}{1-2+14}=\frac{26}{13}\)
\(=2.\)
উত্তর : নির্ণেয় মান \(=2.\)

\(\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta\)
\(=1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\)
\( = 1 + {\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)^2}\)
\(= \frac{{{{\sin }^2}\theta + \cos^2 \theta }}{{{{\sin }^2}\theta }}\)
\(= \frac{1}{{{{\sin }^2}\theta }}\) \(\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\)
\(\therefore cosec ^2\theta = \frac{1}{{{{\sin }^2}\theta }}\)
বা, \(cosec \theta = \frac{1}{{{{\sin }}\theta }}\)
আবার \({\sec ^2}\theta = 1 + {\tan ^2}\theta \)
বা, \({\tan ^2}\theta = {\sec ^2}\theta - 1,\)
বা, \({\tan ^2}\theta = \frac{1}{{{{\cos }^2}\theta }} - 1\)
বা, \({\tan ^2}\theta = \frac{{1 - {{\cos }^2}\theta }}{{{{\cos }^2}\theta }}\)
বা, \(\tan ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\)
বা, \(\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}\)
বা, \(\tan \theta = \frac{{\sin \theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}\)
\(\therefore\) \(\operatorname{cosec} \theta=\frac{1}{\sin \theta}\)
\(\tan \theta=\frac{\sin \theta}{\sqrt{1-\sin ^{2} \theta}}\)

\({cosec ^2}\theta = 1 + {\cot ^2}\theta \)
\( = 1 + {\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)^2}\)
\( = 1 + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\)
\(= \frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\)
\(=\frac{1}{\sin ^{2} \theta}\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\)
\(\therefore\) \( coscec{ ^2} = \frac{1}{{{{\sin }^2}\theta }}\)
বা, \(coscec \theta = \frac{1}{{\sin \theta }}\)
বা, \(coscec \theta = \frac{1}{{\sqrt {1 - {{\cos }^2}\theta } }}\)
আবার, \({\sec ^2} = 1 + {\tan ^2}\theta \)
বা, \({\tan ^2}\theta = {\sec ^2}\theta - 1\)
বা, \({\tan ^2}\theta = \frac{1}{{{{\cos }^2}\theta }} - 1\)
বা, \({\tan ^2}\theta = \frac{{1 - \cos ^2\theta }}{{{{\cos }^2}\theta }}\)
বা, \({\tan ^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\)
বা, \(\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]\)
বা, \(\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}\)
বা, \(\tan \theta = \frac{{\sqrt {1 - \cos^2 \theta } }}{{\cos \theta }}\)
\(\therefore\) \(\operatorname{cosec} \theta=\frac{1}{\sqrt{1-\cos ^{2} \theta}}\)
\(\tan \theta=\frac{\sqrt{1-\cos ^{2} \theta}}{\cos \theta}\)

আমরা জানি, \(\sec ^{2} \theta-\tan ^{2} \theta=1\)
বা, \((\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1\)
বা, \(2(\sec \theta-\tan \theta)=1[\sec \theta+\tan \theta=2]\)
বা, \(\sec \theta-\tan \theta=\frac{1}{2}.\)
\(\therefore(\sec \theta-\tan \theta)\) -এর মান \(=\frac{1}{2}.\)

\({cosec ^2}\theta - {\cot ^2}\theta = 1\)
বা, \(( cosec \theta + \cot \theta )( cose \theta + \cot \theta ) = 1\)
বা, \( (cosec\theta + \cot \theta ) \times \sqrt (2 - 1) = 1\)
বা, \( cosec \theta + \cot \theta = \frac{1}{{\sqrt 2 - 1}}\)
বা, \( cosec \theta + cot \theta = \frac{{\sqrt 2 + 1}}{{{{(\sqrt 2 )}^2} - {{(1)}^2}}}\)
বা, \(cosec \theta + cot \theta =\frac{\sqrt{2}+1}{2-1}\)
বা, \(cosec \theta + cot \theta = \frac{{\sqrt 2 + 1}}{1}\)
বা, \( cosec \theta + cot \theta = \sqrt 2 + 1\)
\(\therefore\)\((cosec \theta + \cot \theta) \)-এর মান \(=\sqrt{2}+1\)

\({\sin ^2} + cos^2 \theta = 1\)
বা, \({(\sin \theta + \cos \theta )^2} - 2\sin \theta \cos \theta = 1\)
বা, \({(1)^2} - 2\sin \theta \cos \theta = 1\) \([\because \sin \theta+\cos \theta=1]\)
বা, \( - 2\sin \theta \cos \theta = 1 - 1\)
বা, \(-2 \sin \theta \cos \theta=0\)
বা, \(\sin \theta \cos \theta = 0\)
\(\therefore \sin \theta \times \cos \theta-\)এর মান \(= 0\)

দেওয়া আছে, \(\tan \theta+\cot \theta=2\)
বা, \(\tan \theta+\frac{1}{\tan \theta}=2\)
বা, \(\frac{\tan ^{2} \theta+1}{\tan \theta}=2\)
বা, \(\tan ^{2} \theta+1=2 \tan \theta\)
বা, \(\tan ^{2} \theta-2 \tan \theta+1=0\)
বা, \((\tan \theta)^{2}-2 \times \tan \theta \times 1+(1)^{2}=0\)
বা, \((\tan \theta-1)^{2}=0\)
বা, \(\tan \theta-1=0\)
বা, \(\tan \theta=1.\)
\(\therefore \cot \theta=\frac{1}{\tan \theta}=\frac{1}{1}=1.\)
\(\therefore \tan \theta-\cot \theta=1-1=0.\)
\(\therefore(\tan \theta-\cot \theta)\)-এর মান \(= 0\)

প্রথম পদ্ধতি :
\(\because \sin \theta-\cos \theta=\frac{7}{13}\) \(\cdots\) \(\cdots\)(i)
\(\sin \theta+\cos \theta=x\) (ধরা যাক) \(\ldots\) \(\ldots\)(ii)
(i) এবং (ii) বর্গ করে এবং যোগ করে,
\((\sin \theta-\cos \theta)^{2}+(\sin \theta+\cos
\theta)^{2}=\frac{49}{169}+x^{2}\)
\(2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\frac{49}{169}+x^{2}\)
বা, \(2(1)-\frac{49}{169}=x^{2}\)\(\left[\sin ^{2} \theta+\cos ^{2} \theta > 1\right]\)
বা, \(\frac{338-49}{169}=x^{2}\)
বা, \(\frac{289}{169}=x^{2}\)
\(\therefore x=\sqrt{\frac{289}{169}}=\frac{17}{13},[\theta\) হল ধনাত্মক
সুক্ষকোণ, অতএব ধনাত্মক মান নিয়ে]
দ্বিতীয় পদ্ধতি :
যেহেতু, \(\sin \theta-\cos \theta=\frac{7}{13}\)
\(\therefore \quad(\sin \theta-\cos
\theta)^{2}=\left(\frac{7}{13}\right)^{2}=\frac{49}{169}\)
\(\therefore \quad \sin ^{2} \theta+\cos ^{2} \theta-2 \sin \theta \cos
\theta=\frac{49}{169}\)
বা, \(1-2 \sin \theta \cos \theta=\frac{49}{169}\)\(\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]\)
বা, \(1-\frac{49}{169}=2 \sin \theta \cos \theta\)
বা, \(\frac{169-49}{169}=2 \sin \theta \cos \theta\)
বা, \(\frac{120}{169}=2 \sin \theta \cos \theta\)
বা, \(\frac{60}{169}=\sin \theta \cos \theta\)
এখন, \((\sin \theta+\cos \theta)^{2}=(\sin \theta-\cos \theta)^{2}+4 \sin \theta
\cos \theta\)
\(=\left(\frac{7}{13}\right)^{2}+4 \times \frac{60}{169}\)
\(=\frac{49}{169}+\frac{240}{169}=\frac{289}{169}\)
\(\therefore(\sin \theta+\cos
\theta)=\sqrt{\frac{289}{169}}=\frac{17}{13},(\because \theta\) ধনাত্মক
সূক্ষকোণ)
\(\therefore \sin \theta+\cos \theta=\frac{17}{13}\) (উত্তর)

\({\sin ^2}\theta + {\cos ^2}\theta = 1\)
বা, \({(\sin \theta + \cos \theta )^2} - 2\sin \theta \cos \theta = 1\)
বা, \({(\sin \theta + \cos \theta )^2} - 2 \times \frac{1}{2} = 1\)\(\left[\because \sin \theta \cos \theta=\frac{1}{2}\right]\)
বা, \((\sin \theta + \cos \theta)^2 = 1 + 1 = 2\)
বা, \(\sin \theta + \cos \theta = \sqrt 2 \)
\(\therefore\) নির্ণেয় \((\sin \theta+\cos \theta)\)-এর মান \(=\sqrt{2}\)

\({\sec ^2}\theta - {\tan ^2}\theta = 1\)
দেওয়া আছে, \( \sec \theta - \tan \theta = \frac{1}{{\sqrt 3 }}\)........ (i)
আমরা জানি, \(\sec ^{2} \theta \ln ^{3} \tan ^{2} \theta=1\)
বা, \((\sec \theta + \tan \theta )(\sec \theta - \tan \theta ) = 1\)
বা, \((\sec \theta + \tan \theta ) \times \frac{1}{{\sqrt 3 }} = 1\)
বা, \(\sec \theta + \tan \theta = \sqrt 3\)...........(ii)
(i)+(ii) করে পাই,
\(\sec \theta-\tan \theta+\sec \theta+\tan \theta=\frac{1}{\sqrt{3}}+\sqrt{3}\)
বা, \(2\sec \theta = \sqrt 3 + \frac{1}{{\sqrt 3 }}\)
বা, \(2\sec \theta = \frac{{3 + 1}}{{\sqrt 3 }}\)
বা, \(2 \sec \theta=\frac{4}{\sqrt{3}}\)
বা, \(\sec \theta = \frac{4}{{\sqrt 3 \times 2}}\)
বা, \(\sec \theta = \frac{2}{{\sqrt 3 }}\)
(ii) নং সমীকরণে \(\sec \theta \)-এর মান বসিয়ে পাই,
\(\sec \theta + \tan \theta = \sqrt 3 \)
বা, \(\tan \theta = \sqrt 3-\frac{2}{{\sqrt 3 }}\)
বা, \(\tan \theta = \frac{{3 - 2}}{{\sqrt 3 }}\)
বা, \(\tan \theta = \frac{1}{{\sqrt 3 }}\)
\(\therefore \sec \theta-\) এর মান \(\frac{2}{\sqrt{3}}\) ও \(\tan \theta\)-এর মান \(=\frac{1}{\sqrt{3}}\)

আমারা জানি, \({cosec ^2}\theta - {\cot ^2}\theta = 1\) দেওয়া আছে \(cosec \theta + \cot \theta = \sqrt 3 \ldots (i)\)
বা, \(\left( {{{cosec }}\theta + \cot \theta } \right)( cosec \theta -\cot \theta) = 1 \)
বা, \( \sqrt 3 (cosec \theta - \cot \theta ) = 1\)
বা, \(cosec \theta - \cot \theta = \frac {1}{\sqrt 3} \ldots \ldots ({\rm{ii}})\)
(i) ও (ii) যোগ করিয়া পাই,
\(\operatorname{cosec} \theta+\cot \theta+\operatorname{cosec} \theta-\cot \theta=\sqrt{3}+\frac{1}{\sqrt{3}}\)
বা, \(2 cosec \theta = \sqrt 3 + \frac{1}{\sqrt 3 }\)
\( = \frac{{3 + 1}}{{\sqrt 3 }} = \frac{4}{{\sqrt 3 }}\)
বা, \(\operatorname{cosec} \theta=\frac{4}{\sqrt{3} \times 2}\)
বা, \(cosec \theta = \frac{2}{{\sqrt 3 }}\)
(i) নং সমীকরণে \(cosec \theta = \frac{2}{{\sqrt 3 }}\) বসিয়ে পাই
\(cosec \theta - \cot \theta = \frac{1}{{\sqrt 3 }}\)
বা, \(\frac{2}{{\sqrt 3 }} - \cot \theta = \frac{1}{{\sqrt 3 }}\)
বা, \( \cot \theta = \frac{{2}}{{\sqrt 3 }}-\frac{1}{{\sqrt 3 }} \)
বা, \(\cot \theta=\frac{2-1}{\sqrt{3}}\)
বা, \(\cot \theta = \frac{1}{{\sqrt 3 }}\)
\(\therefore\) \(\operatorname{cosec} \theta\)-এর মান \(=\frac{2}{\sqrt{3}}\)
\(\cot \theta\)-এর মান \(=\frac{1}{\sqrt{3}}\)

দেওয়া আছে, \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=7\)

বা, \(\frac{\sin \theta+\cos \theta+\sin \theta-\cos \theta}{\sin \theta+\cos \theta-\sin \theta+\cos \theta}=\frac{7+1}{7-1}\)(যোগ-ভাগ প্রক্রিয়া)

বা, \(\frac{2 \sin \theta}{2 \cos \theta}=\frac{8}{6}\)
বা, \(\frac{\sin \theta}{\cos \theta}=\frac{4}{3}\)
বা, \(\tan \theta=\frac{4}{3}=1 \frac{1}{3}.\)

\(\therefore\) \(\tan \theta\)-এর নির্ণেয় মান \(=1 \frac{1}{3}.\)

\(\frac{{cosec \theta + \sin \theta }}{{cosec \theta - \sin \theta }} = \frac{5}{3}\)
বা, \(5cosec \theta - 5\sin \theta = 3cosec \theta + 3\sin \theta \)
বা, \(5cosec \theta - 3cosec \theta = 3\sin \theta + 5\sin \theta \)
বা, \(2cosec \theta = 8\sin \theta \)
বা, \(\frac{1}{{\sin \theta }} = 8\sin \theta \)
বা, \(8{\sin ^2}\theta = 2\)
বা, \({\sin ^2}\theta = \frac{2}{8}\)
বা, \({\sin ^2}\theta = \frac{1}{4}\)
বা, \(\sin \theta = \frac{1}{2}\)

\(\sec \theta+\cos \theta=\frac{5}{2}\)
বা, \((\sec \theta+\cos \theta)^{2}=\left(\frac{5}{2}\right)^{2}\) [বর্গ করে ]
বা, \((\sec \theta-\cos \theta)^{2}+4 \sec \theta \cdot \cos \theta=\frac{25}{4}\)
বা, \((\sec \theta-\cos \theta)^{2}+4 \cdot \frac{1}{\cos \theta} \cdot \cos \theta=\frac{25}{4}\left[\because \sec \theta=\frac{1}{\cos \theta}\right]\)
বা, \((\sec \theta-\cos \theta)^{2}+4=\frac{25}{4}\)
বা, \((\sec \theta-\cos \theta)^{2}=\frac{25}{4}-4\)
বা, \((\sec \theta-\cos \theta)^{2}=\frac{25-16}{4}\)
বা, \((\sec \theta-\cos \theta)^{2}=\frac{9}{4}\)
বা, \(\sec \theta-\cos \theta=\sqrt{\frac{9}{4}}\)
বা, \(\sec \theta-\cos \theta=\frac{3}{2}.\)
\(\therefore (\sec \theta-\cos \theta)\)- এর মান \(=\frac{3}{2}=1 \frac{1}{2}\).

\(5 \sin ^{2} \theta+4 \cos ^{2} \theta=\frac{9}{2}\)
বা, \(10 \sin ^{2} \theta+8 \cos ^{2} \theta=9\)
বা, \(\frac{10 \sin ^{2} \theta}{\cos ^{2} \theta}+8=\frac{9}{\cos ^{2} \theta}\) [উভয়দিকে \(\cos ^{2} \theta\) দ্বারা ভাগ করে]
বা, \(10 \tan ^{2} \theta+8=9\left(1+\tan ^{2} \theta\right)\)
বা, \(10 \tan ^{2} \theta+8=9 \sec ^{2} \theta\)
বা, \(10 \tan ^{2} \theta+8=9+9 \tan ^{2} \theta\)
বা, \(10 \tan ^{2} \theta-9 \tan ^{2} \theta=9-8 \quad\)
বা, \(\tan ^{2} \theta=1\)
বা, \((\tan \theta)^{2}=(\pm 1)^{2}\)
বা, \(\tan \theta=1\) অথবা \(\tan \theta=-1\)
যেহেতু \(\theta\) ধনাত্মক সূক্ষ্মকোণ, \(\therefore \tan \theta=1\) হবে, \(\tan \theta\) এর \(-1\) মান বর্জনীয়।
\(\therefore \quad \tan \theta=1\) এই গ্রহণযােগ্য মান।
উত্তর : \(\tan \theta\) এর মান = 1

\(\tan ^{2} \theta+\cot ^{2} \theta=\frac{10}{3}\)
বা, \((\tan \theta)^{2}+(\cot \theta)^{2}=\frac{10}{3}\)
বা, \((\tan \theta+\cot \theta)^{2}-2 \tan \theta \cdot \cot \theta=\frac{10}{3}\)
বা, \((\tan \theta+\cot \theta)^{2}-2 \tan \theta \cdot \frac{1}{\tan \theta}=\frac{10}{3}\)\(\left[\cot \theta=\frac{1}{\tan \theta}\right]\)
বা, \((\tan \theta+\cot \theta)^{2}-2=\frac{10}{3}\)
বা, \((\tan \theta+\cot \theta)^{2}=\frac{10}{3}+2=\frac{16}{3}\)
বা, \(\tan \theta+\cot \theta=\sqrt{\frac{16}{3}}=\frac{4}{\sqrt{3}}\)
\(\therefore \tan \theta+\cot \theta=\frac{4}{\sqrt{3}}\) \(\ldots (1)\)
আবার, \(\tan ^{2} \theta+\cot ^{2} \theta=\frac{10}{3}\)
বা, \((\tan \theta)^{2}+(\cot \theta)^{2}=\frac{10}{3}\)
বা, \((\tan \theta-\cot \theta)^{2}+2 \tan \theta \times \cot \theta=\frac{10}{3}\)
বা, \((\tan \theta-\cot \theta)^{2}+2 \tan \theta \times \frac{1}{\tan \theta}=\frac{10}{3}\) \(\left[\because \cot \theta=\frac{1}{\tan \theta}\right]\)
বা, \((\tan \theta-\cot \theta)^{2}+2=\frac{10}{3}\)
বা, \((\tan \theta-\cot \theta)^{2}=\frac{10}{3}-2\)
বা, \((\tan \theta-\cot \theta)^{2}=\frac{10-6}{3}\)
বা, \((\tan \theta-\cot \theta)^{2}=\frac{4}{3}\)
বা, \(\tan \theta-\cot \theta=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\)\(\ldots (2)\)
\(\therefore \tan \theta+\cot \theta=\frac{4}{\sqrt{3}}\)
এবং \(\tan \theta-\cot \theta=\frac{2}{\sqrt{3}}.\)
এখন, (1) ও (2) যােগ করে পাই,
\(\tan \theta+\cot \theta+\tan \theta-\cot \theta\)\(=\frac{4}{\sqrt{3}}+\frac{2}{\sqrt{3}}\)
\(2 \tan \theta=\frac{4}{\sqrt{3}}+\frac{2}{\sqrt{3}}\)
বা, \(2 \tan \theta=\frac{4+2}{\sqrt{3}}\)
বা, \(2 \tan \theta=\frac{6}{\sqrt{3}}\)
বা, \(\tan \theta=\frac{6}{\sqrt{3} \times 2}\)
বা, \(\tan \theta=\frac{3}{\sqrt{3}}=\sqrt{3}\)
\(\therefore \tan \theta\) - এর মান \(=\sqrt{3}\).

\({\sec ^2}\theta + {\tan ^2}\theta = \frac{{13}}{{12}}\)
\({\sec ^4}\theta - {\tan ^4}\theta\)
\(= {\left( {{{\sec }^2}\theta } \right)^2} - {\left( {{{\tan }^2}\theta } \right)^2}\)
\( = \left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right)\) \(\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\)
= \(\frac{{13}}{{12}} + 1 = \frac{{13}}{{12}}\)
\(=\frac{13}{12} \times 1\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]=\frac{13}{12}\)
\(\therefore\) নির্ণেয় মান \(=\frac{13}{12}\)

\(\triangle PQR\)-এবং \(\angle Q=90^{\circ}\),
\(\therefore \mathbf{P R}=\) অতিভুজ
এখন, \(\cos P-\cos R\)
\(=\frac{P Q}{P R}-\frac{R Q}{P R}[\operatorname{cosP}\) এবং \(\cos R\)-এর সংজ্ঞানুসারে )
\(=\frac{\mathrm{PQ}-\mathrm{RQ}}{\mathrm{PR}}\)
\(=\frac{1}{\sqrt{5}}[\because P Q-R Q=1\) এবং \(\quad P R=\sqrt{5}]\)
\(\therefore \cos P-\cos R=\frac{1}{\sqrt{5}}\)।

\(\triangle XYZ\) -এ \(\angle \mathrm{Y}=90^{\circ}\).
\(\therefore \mathrm{XZ}=\) অতিভুজ।
এখন, \(\sec X-\tan X\)
\(\frac{ \mathrm{XZ}}{\mathrm{XY}}-\frac{Y\mathrm{X}}{\mathrm{XY}}\) [\(\sec X \) এবং \(\tan X\) -এর সংজ্ঞানুসারে ]
\(=\frac{\mathrm{XZ}-\mathrm{YZ}}{\mathrm{XY}}\)
\(=\frac{2}{2 \sqrt{3}} \quad[\because \mathrm{XZ}-\mathrm{YZ}=2\) এবং \(X Y=2 \sqrt{3}]\)
\(=\frac{1}{\sqrt{3}}.\)
\(\therefore \sec X-\tan X=\frac{1}{\sqrt{3}}.\)

\(x = 2\sin \theta \ldots\) ...(i)
\(y = 3\cos \theta \ldots...(ii)\)
(i) হইতে \(\frac{x}{2} = \sin \theta \ldots....(iii)\)
(ii) হইতে পাই \( \frac{y}{3} = \cos \theta \ldots ....(iv)\)
(iii) ও (iv) নং সমীকরণ বর্গকে যোগ করিয়া পাই,
\(\left(\frac{x}{2}\right)^{2}+\left(\frac{y}{3}\right)^{2}=\sin ^{2} \theta+\cos ^{2} \theta\)
\(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9}\) = \({\sin ^2}\theta + {\cos ^2}\theta\)
বা, \( \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\)

দেওয়া আছে, \(5 x=3 \sec \theta\)
বা, \(\frac{5 x}{3}=\sec \theta\)
বা, \(\left(\frac{5 x}{3}\right)^{2}=\sec ^{2} \theta\)
বা, \(\frac{25 x^{2}}{9}=\sec ^{2} \theta\) \(\ldots(1)\)
আবার, \(y=3 \tan \theta\)
বা, \(\frac{y}{3}=\tan \theta\)
বা, \(\left(\frac{y}{3}\right)^{2}=\tan ^{2} \theta\)
বা, \(\frac{y^{2}}{9}=\tan ^{2} \theta\) \(\ldots(2)\)
এখন, (1) থেকে, (2) বিয়ােগ করে পাই,
\(\frac{25 x^{2}}{9}-\frac{y^{2}}{9}=\sec ^{2} \theta-\tan ^{2} \theta\)
বা, \(\frac{25 x^{2}}{9}-\frac{y^{2}}{9}=1\)\(\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]\)
বা, \(\frac{25 x^{2}-y^{2}}{9}=1\)
বা, \(25 x^{2}-y^{2}=9\)
\(\therefore\) \(\theta\) আপনয়ন করে পাই \(25 x^{2}-y^{2}=9.\)

দেওয়া আছে, \(\sin \alpha=\frac{5}{13}\)
\(\therefore \cos \alpha=\sqrt{1-\sin ^{2} \alpha}\)
\(=\sqrt{1-\left(\frac{5}{13}\right)^{2}}\)\([\because\) sin \alpha =\frac{5}{13}]\)
\(=\sqrt{1-\frac{25}{169}}\)
\(=\sqrt{\frac{169-25}{169}}\)
\(=\sqrt{\frac{144}{169}}=\frac{12}{13}.\)
এখন, \(\tan \alpha+\sec \alpha=\frac{\sin \alpha}{\cos \alpha}+\frac{1}{\cos \alpha}\)
\(=\frac{\frac{5}{13}}{\frac{12}{13}}+\frac{1}{\frac{12}{13}}\)
\(=\frac{5}{13} \times \frac{13}{12}+1 \times \frac{13}{12}\)
\(=\frac{5}{12}+\frac{13}{12}\)
\(=\frac{5+13}{12}=\frac{18}{12}\)
\(=\frac{3}{2}=1 \cdot 5\)
\(\therefore \tan \alpha+\sec \alpha=1.5\) (প্রমাণিত)

দেওয়া আছে, \(\tan \mathrm{A}=\frac{n}{m}\)
ৰা, \(\tan ^{2} \mathrm{~A}=\frac{n^{2}}{m^{2}}\) [ বর্গ কর]
বা, \(1+\tan ^{2} A=1+\frac{n^{2}}{m^{2}}\) [উভয় দিকে 1 যোগ করে]
বা, \(\operatorname{sec}^{2} \mathrm{~A}=\frac{m^{2}+n^{2}}{m^{2}}\)
বা, \(\sec A=\sqrt{\frac{m^{2}+n^{2}}{m^{2}}}\)
বা, \(\sec \mathrm{A} = \frac{\sqrt{m^{2}+n^{2}}}{m}\)
আবার, \(\tan \mathrm{A}=\frac{n}{m}\)
বা, \(\cot \mathrm{A}=\frac{m}{n}\)
বা, \(\cot ^{2} \mathrm{~A}=\frac{m^{2}}{n^{2}}\) [ বর্গ করে]
বা, \(\operatorname{cosec}^{2} \mathrm{~A}-1=\frac{m^{2}}{n^{2}}\) \(\left[\cot ^{2} A=\operatorname{cosec}^{2} A-1\right]\)
বা, \(\operatorname{cosec}^{2} \mathrm{~A}=\frac{m^{2}}{n^{2}}+1\)
বা, \(\operatorname{cosec}^{2} \mathrm{~A}=\frac{m^{2}+n^{2}}{n^{2}}\)
বা, \(\operatorname{cosec} A=\sqrt{\frac{m^{2}+n^{2}}{n^{2}}}\)
বা, \(\operatorname{cosec} \mathrm{A}=\frac{\sqrt{m^{2}+n^{2}}}{n}\)
বা, \(\sin \mathrm{A}=\frac{n}{\sqrt{m^{2}+n^{2}}}\)
\(\therefore \sin \mathrm{A}=\frac{n}{\sqrt{m^{2}+n^{2}}}\)
এবং \(\sec \mathrm{A}=\frac{\sqrt{m^{2}+n^{2}}}{m}\)

যেহেতু \(\cos \theta=\frac{x}{\sqrt{x^{2}+y^{2}}}\)
অতএব \(\sin ^{2} \theta=1-\cos ^{2} \theta\)
\(=1-\left(\frac{x}{\sqrt{x^{2}+y^{2}}}\right)^{2}\)
\(=1-\frac{x^{2}}{x^{2}+y^{2}}=\frac{y^{2}}{x^{2}+y^{2}}\)
বা, \(\sin \theta=\frac{y}{\sqrt{x^{2}+y^{2}}}\) (ধনাত্মক মানটি নিয়ে)
বা, \(x \sin \theta=\frac{x y}{\sqrt{x^{2}+y^{2}}}\) (উভয়দিকে \(x\) গুন করে)
আবার, \(y \cos \theta=\frac{x y}{\sqrt{x^{2}+y^{2}}}\) (উভয়দিকে \(y\) গুন করে)
সুতরাং \(x \sin \theta=y \cos \theta\) (প্রমাণিত)।

\(\cos \theta = \sqrt {1 - \sin \theta } \)
\( = \sqrt {1 - {{\left( {\frac{{{a^2} - {b^2}}}{{{a^2} + {b^2}}}} \right)}^2}} \)
\(= \sqrt {1 - \frac{{{{\left( {{a^2} - {b^2}} \right)}^2}}}{{{{\left( {{a^2} + {b^2}} \right)}^2}}}}\)
\( = \sqrt {\frac{{{{\left( {{a^2} + {b^2}} \right)}^2} - \left( {{a^2} - {b^2}} \right)^2}}{{{{\left( {{a^2} + {b^2}} \right)}^2}}}} \)
\( = \sqrt {\frac{{4{a^2}{b^2}}}{{{{\left( {{a^2} + {b^2}} \right)}^2}}}}\)
\( = {\frac{{2ab}}{{{a^2} + {b^2}}}} \)
বামপক্ষ \(\cot \theta \) \( = \frac{{\cos \theta }}{{\sin \theta }}\)
\(= \frac{{2ab}}{{{a^2} + {b^2}}} \times \frac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}}\)
\(=\frac{2 a b}{a^{2}-b^{2}}\)
\(=\frac{2 a b}{a^{2}-b^{2}}\)
\(= \frac{{2ab}}{{{a^2} - {b^2}}}\) ডানপক্ষ (প্রমাণিত)

দেওয়া আছে, \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}=k\) (ধরো)
\(\therefore \sin \theta=k x\) \(\ldots(1)\)
এবং \(\cos \theta=k y\) \(\ldots (2)\)
আমরা জানি, \(\sin ^{2} \theta+\cos ^{2} \theta=1\)
বা, \((k x)^{2}+(k y)^{2}=1\)
বা, \(k^{2} x^{2}+k^{2} y^{2}=1\)
বা, \(k^{2}\left(x^{2}+y^{2}\right)=1\)
বা, \(k^{2}=\frac{1}{\left(x^{2}+y^{2}\right)}\)
বা, \(k=\sqrt{\frac{1}{\left(x^{2}+y^{2}\right)}}\)
বা, \(k=\frac{1}{\sqrt{x^{2}+y^{2}}}\)
\(\therefore \sin \theta=k x=\frac{1}{\sqrt{x^{2}+y^{2}}} \times x=\frac{x}{\sqrt{x^{2}+y^{2}}}\) \(\ldots (3)\)
এবং \(\cos \theta=k y=\frac{1}{\sqrt{x^{2}+y^{2}}} \times y=\frac{y}{\sqrt{x^{2}+y^{2}}}\) \(\ldots(4)\)
\(\therefore \sin \theta-\cos \theta=\frac{x}{\sqrt{x^{2}+y^{2}}}-\frac{y}{\sqrt{x^{2}+y^{2}}}=\frac{x-y}{\sqrt{x^{2}+y^{2}}}\).
\(\therefore \sin \theta-\cos \theta=\frac{x-y}{\sqrt{x^{2}+y^{2}}}\) (প্রমাণিত)

\(\because\left(1+4 x^{2}\right) \cos A=4 x\)
\(\therefore \cos A=\frac{4 x}{1+4 x^{2}}\)
\(\therefore \sin ^{2} A=1-\cos ^{2} A=1-\frac{(4 x)^{2}}{\left(1+4 x^{2}\right)^{2}}\)
\(=1-\frac{16 x^{2}}{\left(1+4 x^{2}\right)^{2}}=\frac{\left(1+4 x^{2}\right)^{2}-4.4 x^{2} \cdot 1}{\left(1+4 x^{2}\right)^{2}}=\frac{\left(1-4 x^{2}\right)^{2}}{\left(1+4 x^{2}\right)^{2}} .\)
\(\therefore \sin A=\sqrt{\frac{\left(1-4 x^{2}\right)^{2}}{\left(1+4 x^{2}\right)^{2}}}\)
\(\therefore \sin A=\frac{1-4 x^{2}}{1+4 x^{2}} \therefore \operatorname{cosec} A=\frac{1+4 x^{2}}{1-4 x^{2}}\)
\(\cot A=\frac{\cos A}{\sin A}=\frac{\frac{4 x}{1+4 x^{2}}}{\frac{1-4 x^{2}}{1+4 x^{2}}}=\frac{4 x}{1-4 x^{2}}\)
\(\therefore \operatorname{cosec} A+\cot A=\frac{1+4 x^{2}}{1-4 x^{2}}+\frac{4 x}{1-4 x^{2}}=\frac{1+4 x^{2}+4 x}{1-4 x^{2}}\)
\(=\frac{(1)^{2}+2 \times 1 \times 2 x+(2 x)^{2}}{(1)^{2}-(2 x)^{2}}\)
\(=\frac{(1+2 x)^{2}}{(1+2 x)(1-2 x)}=\frac{1+2 x}{1-2 x}\)(প্রমাণিত)

দেওয়া আছে, \(x=a \sin \theta\)
বা, \(\frac{x}{a}=\sin \theta\)
বা, \(\frac{x^{2}}{a^{2}}=\sin ^{2} \theta\) (উভয়দিকে বর্গ করে)
বা, \(\frac{a^{2}}{x^{2}}=\frac{1}{\sin ^{2} \theta}\)
বা, \(\frac{a^{2}}{x^{2}}=\operatorname{cosec}^{2} \theta\) \(\ldots(1)\)
আবার, \(y=b \tan \theta\)
বা, \(\frac{y}{b}=\tan \theta\)
বা, \(\frac{b}{y}=\cot \theta\) (উভয়দিকে বর্গ করে)
বা, \(\cot ^{2} \theta=\frac{b^{2}}{y^{2}}\) \(\ldots(2)\)
এখন, (1) থেকে (2) বিয়ােগ করে পাই,
\(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}\)
বা, \(1=\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}\)
\(\therefore \frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}=1\) (প্রমাণিত)

দেওয়া আছে, \(\sin \theta+\sin ^{2} \theta=1\)
বা, \(\sin \theta=1-\sin ^{2} \theta\)
বা, \(\sin \theta=\cos ^{2} \theta\) \(\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right]\)
বা, \(\sin ^{2} \theta=\cos ^{4} \theta\) [বর্গ করে]
বা, \(1-\cos ^{2} \theta=\cos ^{4} \theta\)
বা, \(1=\cos ^{2} \theta+\cos ^{4} \theta\)
বা, \(\cos ^{2} \theta+\cos ^{4} \theta=1.\)
\(\therefore \cos ^{2} \theta+\cos ^{4} \theta=1\) (প্রমাণিত)

দেওয়া আছে, \(3 x=\operatorname{cosec} \alpha\)
বা, \((3 x)^{2}=\operatorname{cosec}^{2} \alpha\) (উভয়দিকে বর্গ করে)
বা, \(9 x^{2}=\operatorname{cosec}^{2} \alpha\) \(\ldots (1)\)
আবার, \(\frac{3}{x}=\cot \alpha\)
বা, \(\left(\frac{3}{x}\right)^{2}=(\cot \alpha)^{2}\) (উভয়দিকে বর্গ করে)
বা, \(\frac{9}{x^{2}}=\cot ^{2} \alpha\) \(\ldots (2)\)
আমরা জানি, \(\operatorname{cosec}^{2} \alpha-\cot ^{2} \alpha=1\)
বা, \(9 x^{2}-\frac{9}{x^{2}}=1\)
বা, \(9\left(x^{2}-\frac{1}{x^{2}}\right)=1\)
বা, \(x^{2}-\frac{1}{x^{2}}=\frac{1}{9}\)
বা, \(3\left(x^{2}-\frac{1}{x^{2}}\right)=3 \times \frac{1}{9}\) (উভয়দিকে 3 গুন করে)
বা, \(3\left(x^{2}-\frac{1}{x^{2}}\right)=\frac{1}{3}\)
\(\therefore\) (c) উত্তরটি সঠিক।

\(2x = \sec A, \frac{2}{x} = \tan A \)
উভয় পক্ষকে বর্গ এবং বিয়োগ করিয়া পাই।
\({\sec ^2}A - {\tan ^2}A = {(2x)^2} - {\left( {\frac{2}{x}} \right)^2}\)
বা, \(1 = 4{x^2} - \frac{4}{{{x^2}}}\)
বা, \(4\left( {{x^2} - \frac{1}{{{x^2}}}} \right) = 1\)
বা, \(2 \times 2\left(x^{2}-\frac{1}{x^{2}}\right)=1\)
বা, \(2\left( {{x^2} - \frac{1}{{{x^2}}}} \right) = \frac{1}{2}\)
\((a)\frac{1}{2}\)

দেওয়া আছে, \(\tan \alpha+\cot \alpha=2\)
বা, \(\tan \alpha+\frac{1}{\tan \alpha}=2\)
বা, \(\frac{\tan ^{2} \alpha+1}{\tan \alpha}=2\)
বা, \(\tan ^{2} \alpha+1=2 \tan \alpha\)
বা, \(\tan ^{2} \alpha-2 \tan \alpha+1=0\)
বা, \((\tan \alpha-1)^{2}=0\)
বা, \(\tan \alpha-1=0\)
বা, \(\tan \alpha=1\)
\(\therefore \cos \alpha=\frac{1}{\tan \alpha}=\frac{1}{1}=1\)
\(\therefore \tan ^{13} \alpha+\cot ^{13} \alpha=(1)^{13}+(1)^{13}\)
\(=1+1=2.\)
\(\therefore\) (c) উত্তরটি সঠিক।

\(\sin \theta - \cos \theta = 0\)
বা, \(\sin \theta = \cos \theta \)
বা, \(\sin \theta = \sin \left( {{{90}^\circ } - \theta } \right)\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
বা, \(\theta = {90^\circ }- \theta \)
বা, \(\theta+\theta=90^{\circ}\)
বা, \(20=90^{\circ}\)
বা, \(\theta=\frac{90^{\circ}}{2}\)
বা, \(\theta = 45\)
প্রদত্ত, \(\sec \theta + cosec \theta = x\)
বা, \(\sec 45^\circ + cosec {45^\circ } = x\) \(\left[\because \theta=45^{\circ}\right]\)
বা, \(\sqrt 2 + \sqrt 2 = x\)
বা, \(x = 2\sqrt 2 \)
\({\rm{ (d) }}2\sqrt 2 \)

\(2 \cos 3 \theta=1\)
\(\Rightarrow \cos 3 \theta=\frac{1}{2}=\cos 60^{\circ}\)
\(\Rightarrow 3\theta=60^{\circ}\)
\(\Rightarrow \theta=20^{\circ}\)

সত্য, কারণ,
\(\sec ^{2} \alpha+\cos ^{2} \alpha\)
\(=(\sec \alpha-\cos \alpha)^{2}+2 \sec \alpha \times \cos \alpha\).
\(=(\sec \alpha-\cos \alpha)^{2}+2 \cdot \frac{1}{\cos \alpha} \times \cos \alpha\)
\(=(\sec \alpha-\cos \alpha)^{2}+2\)
এখন, \((\sec \alpha-\cos \alpha)^{2}\)-এর ক্ষুদ্রতম মান হল \(0\).
\(\therefore\left(\sec ^{2} \alpha+\cos ^{2} \alpha\right)\) এর ক্ষুদ্রতম মান \(=0+2=2\)
\(\therefore\) প্রদত্ত বিবৃতিটি সত্য।

মিথ্যা; কারণ,
\(\cos 0^{\circ} \times \operatorname{cos} 1^{\circ} \times \cos 2^{\circ} \times \cos 3^{\circ} \times \ldots \times \cos 90^{\circ}\)
\(=1 \times \cos 1^{\circ} \times \cos 2^{\circ} \times \cos 3^{\circ}\times \ldots \ldots \times0\)
[\(\because \cos ^{\circ}=1\) এবং \(\cos 9 \theta^{\circ} = \theta\).]
\(=0\).
\(\therefore\) প্রদত্ত বিবৃতিটি মিথ্যা।

4; কারণ
\(\frac{4}{\sec ^{2} \theta}+\frac{1}{1+\cot ^{2} \theta}+3 \sin ^{2} \theta.\)
\(=4 \cos ^{2} \theta+\frac{1}{\operatorname{cosec}^{2} \theta}+3 \sin ^{2} \theta.\)
\(=4 \cos ^{2} \theta+\sin ^{2} \theta+3 \sin ^{2} \theta.\)
\(=4 \cos ^{2} \theta+4 \sin ^{2} \theta.\)
\(=4\left(\cos ^{2} \theta+\sin ^{2} \theta\right).\)
\(=4 \times 1=4.\)

সমাধান : \(\sin \left( {\theta - {{30}^\circ }} \right) = \frac{1}{2}\)
বা, \(\sin \left( {\theta - {{30}^\circ }} \right) = \sin {30^\circ }\)
বা, \(\theta-30^{\circ}=30^{\circ}\)
বা, \(0=30^{\circ}+30^{\circ}\)
বা, \(\theta = {60^\circ }\)
\(\therefore \cos \theta\)
=\(\cos {60^\circ } = \frac{1}{2}\)

\(\frac{1}{2}\); কারণ, \(\cos ^{4} \theta-\sin ^{4} \theta\)
\(=\left(\cos ^{2} \theta\right)^{2}-\left(\sin ^{2} \theta\right)^{2}\)
\(=\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)
\(=1 \times\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)\(\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]\)
\(=\cos ^{2} \theta-\sin ^{2} \theta\)
\(=\frac{1}{2}\left[\because \cos ^{2} \theta-\sin ^{2} 0=\frac{1}{2}\right]\)

\(r \cos \theta=2 \sqrt{3}\)
বা, \((\operatorname{rrcos} \theta)^{2}=(2 \sqrt{3})^{2}\)
বা, \(r^{2} \cos ^{2} \theta=12\) [বর্গ করে] \(\ldots(1)\)
আবার, \(r \sin \theta=2\)
বা, \((r \sin \theta)^{2}=(2)^{2}\) বা, \(r^{2} \sin ^{2} \theta=4\) [বর্গ করে] \(\ldots(2)\)
এখন, (1) ও (2) যােগ করে পাই,
\(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=16\)
বা, \(r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=16\)
বা, \(r^{2} \times \cdot (1)=16\)
বা, \(x^{2}=16\)
বা, \(r=\sqrt{16}=4\)
আবার, \(\frac{r \cos \theta}{r \sin \theta}=\frac{2 \sqrt{3}}{2}\)
বা, \(\cot \theta=\sqrt{3}\)
বা, \(\cot \theta=\cot 30^{\circ}\left(\because 0^{\circ}< \theta < 90^{\circ}\right)\)
\(\Rightarrow \theta=30^{\circ}\)
\(\therefore r=4\), \(\theta=30^{\circ}\)
এবং \(\theta=30^{\circ}\)

দেওয়া আছে, \(0^{\circ} \leq \mathrm{A} \leq 90^{\circ}\) এবং \(0^{\circ} \leq \mathrm{B} \leq 90^{\circ}\)
আবার, \(\sin A+\sin B=2.\)
\(\therefore\) স্বাভাবিকভাবেই \(\sin A=1\) এবং \(\sin B=1\). [\(\because\) \(\sin A\) এবং \(\sin B\)-এর বৃহত্তম মান \(= 1\)]
এখন, \(\sin A=1 \Rightarrow \sin A=\sin 90^{\circ}\)
\(\Rightarrow \mathrm{A}=90^{\circ}\)
আবার, \(\sin B=1 \Rightarrow \sin B=\sin 90^{\circ}\)
\(\Rightarrow \mathbf{B}=90^{\circ}\)
\(\therefore \cos A+\cos B=\cos 90^{\circ}+\cos 90^{\circ}\)\(\left[\because A=90^{\circ}, B=90^{\circ}\right]\)
\(=0+0=0\)
\(\therefore(\cos A+\cos B)\)-এর নির্ণেয় মান \(= 0\).

\(9{\tan ^2}\theta + 4{\cot ^2}\theta \)
\( = {(3\tan \theta )^2} + {(2\cot \theta )^2}\)
\( = {(3\tan \theta - 2\cot \theta )^2} -2.3\tan \theta \times 2\cot \theta \)
\(=(3 \tan \theta-2 \cot \theta)^{2}-12 \tan \theta \times \frac{1}{\tan \theta}\)
\( = (3\tan \theta - 2\cot \theta )^2 + 12\)
যে-কোনো পূর্ণবর্গ সংখ্যামালার সর্বনিন্ম মান 0 (শূন্য)
সুতরাং \(9{\tan ^2}\theta + 4{\cot ^2}\) এর নূন্যতম মান 12

\(\sin ^{6} \alpha+\cos ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha.\)
\(=\left(\sin ^{2} \alpha\right)^{3}+\left(\cos ^{2} \alpha\right)^{3}+3 \sin ^{2} \alpha \cos ^{2} \alpha .\)
\(=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{3}-3 \sin ^{2} \alpha \cos ^{2} \alpha\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+3 \sin ^{2} \alpha \cos ^{2} \alpha.\)
\(=(1)^{3}-3 \sin ^{2} \alpha \cos ^{2} \alpha \times 1+3 \sin ^{2} \alpha \cos ^{2} \alpha.\)
\(=1-3 \sin ^{2} \alpha \cos ^{2} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha=1.\)
\(\therefore\) প্রদত্ত রাশির নিয়ে মান \(= 1\).

\(\operatorname{cosec}^{2} \theta=2 \cot \theta\)
বা, \(1+\cot ^{2} \theta=2 \cot \theta\)
বা, \(1+\cot ^{2} \theta-2 \cot \theta=0\)
বা, \(1-2 \cot \theta+\cot ^{2} \theta=0\)
বা, \((1-\cot \theta)^{2}=0\)
বা, \(1-\cot \theta=0\)
বা, \(\cot \theta=1=\cot 45^{\circ} \therefore \theta=45^{\circ} .\)
\(\therefore \theta\)-এর মান \(=45^{\circ}\)