সমাধান : চিত্রে মনে করি AB বাড়িটির জানালা। ভূমি থেকে যার উচ্চতা \(x\) মিটার।
Sin \({60^\circ } = \frac{{AB}}{{AC}}\) [সংজ্ঞানুসারে]
বা, \(\frac{{\sqrt 3 }}{2} = \frac{x}{{2\sqrt 3 }}\)
বা, \(2 x=2 \sqrt{3} \times \sqrt{3}\)
বা, \(2 x=2 \times 3\)
বা, \(2x = 6\)
বা, \(x=\frac{6}{2}\)
বা, \({{x = 3}}\)
\(\therefore\) জানালাটি ভূমি থেকে 3 মিটার উচ্চে আছে।

\(ABC\) সমকোণী ত্রিভুজে \(\angle B=90^{\circ}.\)
\(\therefore \angle \mathrm{ACB}+\angle \mathrm{BAC}=90^{\circ}\) \(\ldots (1)\)
এখন, \(\tan \theta\)-এর সংজ্ঞানুসারে,
\(\tan \angle \mathrm{ACB}=\frac{\mathrm{AB}}{\mathrm{BC}}\) \(\left[\because \tan \theta=\frac{\text { লম্ব }}{\text { ভূমি }}\right]\)
\(=\frac{8 \sqrt{3}}{8}=\sqrt{3}=\tan 60^{\circ}\)
\(\therefore \angle \mathrm{ACB}=60^{\circ}\)
(l) থেকে পাই, \(\angle B A C=90^{\circ}-\angle A CB\)
\(=90^{\circ}-60^{\circ}=30^{\circ}\)
\(\therefore\) নির্ণেয় মান \(\angle \mathrm{ACB}=60^{\circ}\) এবং \(\angle \mathrm{BAC}=30^{\circ}\).

সমাধান :
\(\angle A = {30^\circ },\angle B = 90^\circ ,\angle C\) = \(180^{\circ}-(\angle B+\angle A)\) \(= 180 - \left( {{{90}^\circ } + {{30}^\circ }} \right) = {60^\circ }\)
\(\therefore\) \(\sin {60^\circ } = \frac{{AB}}{{AC}}\) [সংজ্ঞানুসারে]
বা, \(\frac{{\sqrt 3 }}{2} = \frac{{AB}}{{20}}\)
বা, \(2AB = 20\sqrt 3 \)
\(AB = 10\sqrt 3 \) মিটার
আবার \(\cos {60^\circ } = \frac{{{\rm{BC}}}}{{{\rm{AC}}}}\) [সংজ্ঞানুসারে]
বা, \(\frac{1}{2} = \frac{{BC}}{{20}}\)
বা, \(B C=\frac{20}{2}\)
বা, \({\rm{B C = 10}}\) মিটার
\(\therefore B C=10\) মিটার
\(A B=10 \sqrt{3}\) মিটার

\(PQR\) সমকোণী ত্রিভুজে, \(\angle Q=1\) সমকোণ এবং \(\angle \mathrm{R}=45^{\circ}.\)
\(\therefore \sin 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{PR}}.\)[সংজ্ঞানুসারে]
বা, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{PQ}}{\mathrm{PR}}\)
বা, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{PQ}}{3 \sqrt{2}}\)
বা, \(P Q=\frac{3 \sqrt{2}}{\sqrt{2}}\)
বা, \(\mathrm{PQ}=3\)
আবার, \(\cos 45^{\circ}=\frac{\mathrm{QR}}{\mathrm{PR}}\) [সংজ্ঞানুসারে]
বা, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{QR}}{\mathrm{PR}}\)
বা, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{QR}}{3 \sqrt{2}}\)
বা, \(Q R=\frac{3 \sqrt{2}}{\sqrt{2}}\)
বা, \(\mathrm{QR}=3.\)
\(\therefore\) \(PQ\) ও \(QR\) উভয় বাহুর দৈর্ঘ্য \(3\) সেমি।

=\({\left( {\frac{1}{{\sqrt 2 }}} \right)^2} - {\left( {\frac{2}{{\sqrt 3 }}} \right)^2} + {\left( {\frac{2}{{\sqrt 3 }}} \right)^2}\)
\( = \frac{1}{2} - \frac{4}{3} + \frac{4}{3}\)
\(= \frac{1}{2}\)
\(\therefore\) নির্ণেয় মান \(= \frac{1}{2}\)

= \({(\sqrt 2 )^2} - {(1)^2} - {\left( {\frac{1}{2}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}\)
\(= 2 - 1 - \frac{1}{4} - \frac{3}{4} \)
\(= \frac{{8 - 4 - 1 - 3}}{4}\)
\(= \frac{{8 - 8}}{4}\)
\(= \frac{0}{4}\)
\( = 0\)
\(\therefore\) নির্ণেয় মান \( = 0\)

প্রদত্ত রাশিমালা
\(=3 \tan ^{2} 45^{\circ}-\sin ^{2} 60^{\circ}-\frac{1}{3} \cot ^{2} 30^{\circ}-\frac{1}{8} \sec ^{2} 45^{\circ}\)
\(=3 \times(1)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{1}{3}(\sqrt{3})^{2}-\frac{1}{8}(\sqrt{2})^{2}\)
\(=3-\frac{3}{4}-\frac{3}{3}-\frac{1}{8} \times 2\)
\(=3-\frac{3}{4}-1-\frac{1}{4}\)
\(=\frac{12-3-4-1}{4}\)
\(=\frac{12-8}{4}\)
\(=\frac{4}{4}=1\)
\(\therefore\) নির্ণেয় মান \(= 1\)

\(\frac{4}{3} \cot ^{2} 30^{\circ}+3 \sin ^{2} 60^{\circ}-2 \operatorname{cosec}^{2} 60^{\circ}-\frac{3}{4} \tan ^{2} 30^{\circ}.\)
\(=\frac{4}{3} \times(\sqrt{3})^{2}+3 \times\left(\frac{\sqrt{3}}{2}\right)^{2}-2 \times\left(\frac{2}{\sqrt{3}}\right)^{2}-\frac{3}{4} \times\left(\frac{1}{\sqrt{3}}\right)^{2}\)
\(=\frac{4}{3} \times 3+3 \times \frac{3}{4}-2 \times \frac{4}{3}-\frac{3}{4} \times \frac{1}{3}.\)
\(=4+\frac{9}{4}-\frac{8}{3}-\frac{1}{4}=\frac{48+27-32-3}{12}\)
\(=\frac{75-35}{12}=\frac{40}{12}=\frac{10}{3}=3 \frac{1}{3}\)
\(\therefore\) নির্ণেয় মান \(=3 \frac{1}{3}\).

= \(\frac{{\frac{1}{3} \times \frac{{\sqrt 3 }}{2}}}{{\frac{1}{2} \times \frac{1}{{\sqrt 2 }}}} + \frac{\sqrt 3 }{\frac{{\sqrt 3 }}{2}}\)
\(= \frac{{\sqrt 3 }}{6} \times \frac{{2\sqrt 2 }}{1} + \sqrt 3 \times \frac{2}{{\sqrt 3 }}\)
\(= \frac{{\sqrt 6 }}{3} + 2\)
\(= \frac{{\sqrt 6 + 6}}{3}\) \(\therefore\) \(\therefore\) নির্ণেয় মান \(1=\frac{\sqrt{6}+6}{3}\)

=\({(\sqrt 3 )^2} - 2 \times {\left( {\frac{1}{2}} \right)^2} - \frac{3}{4}{(\sqrt 2 )^2} - 4{\left( {\frac{1}{2}} \right)^2}\)
\( = 3 - 2 \times \frac{1}{4} - \frac{3}{4} \times 2 - 4 \times \frac{1}{4}\)
\(= 3 - \frac{1}{2} - \frac{3}{2} – 1\)
\(= \frac{{6 - 1 - 3-2}}{2}\)
\(= \frac{{6 - 6}}{2}\)
\( = 0\)
\(\therefore\) নির্ণেয় মান \( = 0\)

\(={(2)^2} - {(\sqrt 3 )^2} - \frac{{2 \times \frac{1}{{\sqrt 3 }} \times \frac{2}{{\sqrt 3 }}}}{{1 + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}\)
\(= 4 - 3 - \frac{{\frac{4}{3}}}{{\frac{4}{3}}}\)
\(= 4 - 3 - 1 \)
\(= 4 – 4\)
\( = 0\)
\(\therefore\) নির্ণেয় মান \( = 0\)

\(=\frac{{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}} \times \sqrt{3}}+\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}\)
\(=\frac{\frac{3-1}{\sqrt{3}}}{1+1}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\)
\(=\frac{\frac{2}{\sqrt{3}}}{2}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\)
\(= \frac{2}{{\sqrt 3 }} \times \frac{1}{2} + \frac{{\sqrt 3 }}{4} + \frac{{\sqrt 3 }}{4}\)
\(= \frac{1}{{\sqrt 3 }} + \frac{{\sqrt 3 }}{4} + \frac{{\sqrt 3 }}{4}\)
\(= \frac{{4 + 3 + 3}}{{4\sqrt 3 }}\)
\(= \frac{{10}}{{4\sqrt 3 }}\)
\(= \frac{5}{{2\sqrt 3 }}\)
\(\therefore\) নির্ণেয় মান \(=\frac{5}{2 \sqrt{3}}\)

= \(\frac{{1 - {{\left( {\frac{1}{2}} \right)}^2}}}{{1 + {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} \times \frac{{{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}{{1 - 0}} \div \left( {\frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 3 }}} \right)\)
\( = \frac{{1 - \frac{1}{4}}}{{1 + \frac{1}{2}}} \times \frac{{\frac{1}{4} + \frac{3}{4}}}{1} \div \frac{1}{2}\)
\(=\frac{\frac{4-1}{4}}{\frac{2+1}{2}} \times \frac{\frac{1+3}{4}}{1} \div \frac{1}{2}\)
\(=\frac{\frac{3}{4}}{3 / 2} \times \frac{\frac{4}{4}}{1} \div \frac{1}{2}\)
\(= \frac{3}{4} \times \frac{2}{3} \times \frac{4}{4} \times 1 \div \frac{1}{2}\)
\(= \frac{1}{2} \div \frac{1}{2}\)
\(=\frac{1}{2} \times \frac{2}{1}\)
\(= 1\)
\(\therefore\) নির্ণেয় মান \(= 1\)

বামপক্ষ =\({\sin ^2}{45^\circ } + {\cos ^2}{45^\circ }\)
\( = {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}\)
\(= \frac{1}{2} + \frac{1}{2} = 1 = \) ডানপক্ষ (প্রমাণিত)

বামপক্ষ \(=\cos 60^{\circ}=\frac{1}{2}\)
ডানপক্ষ \(=\cos ^{2} 30^{\circ}-\sin ^{2} 30^{\circ}\)
\(=\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}=\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}\)
\(=\frac{2}{4}=\frac{1}{2} .\)
\(\therefore\) বামপক্ষ = ডানপক্ষ। (প্রমাণিত)

সমাধান : বামপক্ষ = \(\frac{{2\tan {{30}^\circ }}}{{1 - {{\tan }^2}{{30}^\circ }}}\)
\(= \frac{{2 \times \left( {\frac{1}{{\sqrt 3 }}} \right)}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}\)
\(=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\)
\(=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}\)
\( = \)\(\frac{\frac{2}{\sqrt 3 }}{{\frac{2}{3}}}\)
\( = \frac{2}{{\sqrt 3 }} \times \frac{3}{2} =\frac{{\sqrt 3 \times \sqrt 3 }}{{\sqrt 3 }}\)
\(= \sqrt 3 = \) ডানপক্ষ (প্রমাণিত)

বামপক্ষ \(=\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}\)
\(=\sqrt{\frac{\left(1+\cos 30^{\circ}\right)\left(1+\cos 30^{\circ}\right)}{\left(1-\cos 30^{\circ}\right)\left(1+\cos 30^{\circ}\right)}}\)
\(=\sqrt{\frac{\left(1+\cos 30^{\circ}\right)^{2}}{1-\cos ^{2} 30^{\circ}}}=\sqrt{\frac{\left(1+\cos 30^{\circ}\right)^{2}}{\sin ^{2} 30^{\circ}}}\) \(\left[\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right]\)
\(=\frac{1+\cos 30^{\circ}}{\sin 30^{\circ}}=\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{1}{2}}=\frac{2+\sqrt{3}}{2} \times \frac{2}{1}=2+\sqrt{3}\)
ডানপক্ষ \(=\tan 60^{\circ}+\sec 60^{\circ}\) \(=\sqrt{3}+2\)
\(\therefore\) বামপক্ষ= ডানপক্ষ (প্রমাণিত )

বামপক্ষ = \(\frac{{2 {{\tan }^2}{{30}^\circ }}}{{1 - {{\tan }^2}{{30}^\circ }}} + {\sec ^2}{45^\circ } - {\cot ^2}{45^\circ }\)
\(=\frac{{2 \times {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{1 - {{\left( {\frac{1}{\sqrt 3}} \right)}^2}}} + {(\sqrt 2 )^2} - {(1)^2}\)
\(=\frac{\frac{2}{3}}{1-\frac{1}{3}}+2-1\)
\(=\frac{\frac{2}{3}}{\frac{3-1}{3}}+2-1\)
\(=\frac{{\frac{2}{3}}}{{\frac{2}{3}}} + 2 – 1\)
\( = 1 + 2 - 1\)
\( = 2\)
ডানপক্ষ \(\sec {60^\circ } = 2\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)

বামপক্ষ \(=\tan ^{2} \frac{\pi}{4} \cdot \sin \frac{\pi}{3}, \tan \frac{\pi}{6} \tan ^{2} \frac{\pi}{3}\)
\(=\tan ^{2} 45^{\circ} \cdot \sin 60^{\circ} \cdot \tan 30^{\circ} \tan ^{2} 60^{\circ}\)
\(=(1)^{2} \times \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} \times (\sqrt{3})^{2}=(1)^{2} \times \frac{1}{2} \times 3=\frac{3}{2}\)(প্রমাণিত )

বামপক্ষ \(=\sin \frac{\pi}{3} \tan \frac{\pi}{6}+\sin \frac{\pi}{2} \cos \frac{\pi}{3}\)
\(=\sin 60^{\circ} \tan 30^{\circ}+\sin 90^{\circ} \cos 60^{\circ}\left[\because \pi=180^{\circ}\right]\)
\(=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}+1 \times \frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1\)
ডানপক্ষ \(=2 \sin ^{2} \frac{\pi}{4}\)
\(=2 \times \sin ^{2} 45^{\circ}\)
\(=2 \times\left(\frac{1}{\sqrt{2}}\right)^{2}\)
\(=2 \times \frac{1}{2}=1.\)
\(\therefore\) বামপক্ষ = ডানপক্ষ। (প্রমাণিত)

\(x \sin 45^{\circ} \cos 45^{\circ} \tan 60^{\circ}=\tan ^{2} 45^{\circ}-\cos 60^{\circ}.\)
বা, \(x \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \sqrt{3}=(1)^{2}-\frac{1}{2}\)
বা, \(x \times \frac{\sqrt{3}}{2}=1-\frac{1}{2}\)
বা, \(x \times \frac{\sqrt{3}}{2}=\frac{2-1}{2}\)
বা, \(x \times \frac{\sqrt{3}}{2}=\frac{1}{2}\)
বা, \(x=\frac{1}{2} \times \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)
\(\therefore\) \(x\)-এর নির্ণেয় মান \(=\frac{1}{\sqrt{3}}\).

ত্রিকমিতিক কোণের মান বসিয়ে,
\(x\left(\frac{\sqrt{3}}{2}\right)^{2} \cdot \frac{\sqrt{3}}{2}=\frac{(1)^{2}
\cdot 2}{\frac{2}{\sqrt{3}}}\)
বা, \(x \frac{3}{4} \times \frac{\sqrt{3}}{2}=\frac{1 \times 2 \times \sqrt{3}}{2}\)
বা, \(x=\frac{1 \times 2 \times \sqrt{3} \times 4 \times 2}{2 \times 3 \times
\sqrt{3}}=\frac{8}{3}=2 \frac{2}{3}\) (উত্তর)

\(x^{2}=\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}\)
বা, \(x^{2}=\left(\frac{1}{2}\right)^{2}+4 \times(1)^{2}-(2)^{2}\)
বা, \(x^{2}=\frac{1}{4}+4-4\)
বা, \(x^{2}=\frac{1}{4}\)
বা, \(x=\pm \frac{1}{2}\)
\(\therefore\) \(x\)-এর নির্ণেয় মান \(=\pm \frac{1}{2}\).

দেওয়া আছে, \(x \tan 30^{\circ}+y \cot 60^{\circ}=0\)
বা, \(x \times \frac{1}{\sqrt{3}}+y \times \frac{1}{\sqrt{3}}=0\)
বা, \(x + y = 0\) \(\ldots(1)\) \([\sqrt{3}\) দ্বারা গুণ করে]
আবার, \(2 x-y \tan 45^{\circ}=1\)
বা, \(2 x-y \times 1=1\)
বা, \(2 x-y=1 \ldots (2) \)
এখন, (1) ও (2) যােগ করে পাই,
\(3x = 1\)
বা, \(x=\frac{1}{3}\)
(1) থেকে পাই, \(\frac{1}{3}+y=0\)
বা, \(y=-\frac{1}{3}\)
\(\therefore\) নির্ণেয় মান \(x=\frac{1}{3}\) এবং \(y=-\frac{1}{3}\)

(i) বামপক্ষ = \(\sin (A + B)\)
\( = \sin {\left( {{{45}^\circ } + {{45}^\circ }} \right) } =\sin 90^\circ= 1\)
ডানপক্ষ = \(\sin {45^\circ } \cos {45^\circ } + \cos {45^\circ }\sin {45^\circ }\)
\( = \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }}\)
\(= \frac{1}{2} + \frac{1}{2}\)
\( = \frac{2}{2} = 1\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)
(ii) বামপক্ষ = \(=\cos (A+B)\)
\(=\cos \left(45^{\circ}+45^{\circ}\right)\)
\(=\cos 90^{\circ}=0\)
ডানপক্ষ \(=\cos A \cos B-\sin A \sin B\)
\(=\cos 45^{\circ} \cos 45^{\circ}-\sin 45^{\circ} \sin 45^{\circ}\)
\(=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\)
\(=\frac{1}{2}-\frac{1}{2}=0\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)

এখানে, \(\angle \mathrm{ADB}=90^{\circ}\),
\(\therefore\) \(AB =\) অতিভুজ
\(\therefore \tan \angle \mathrm{ABD}=\frac{\mathrm{AD}}{\mathrm{BD}}\) [সংজ্ঞানুসারে] \(\ldots (1)\)
আবার, \(\cot \angle B A D=\frac{A D}{B D}\) [সংজ্ঞানুসারে] \(\ldots(2)\)
(1) ও (2) থেকে আমরা পাই, \(\tan \angle \mathrm{ABD}=\cot \angle \mathrm{BAD}\).
\(\therefore \tan \angle A B D=\cot \angle B A D\). (প্রমাণিত)

\(\triangle \mathrm{ABC}-এ \mathrm{AB}=\mathrm{AC}\),
\(\therefore \angle \mathrm{ABC}=\angle \mathrm{ACB}=45^{\circ} \quad\left[\because \angle \mathrm{BAC}=90^{\circ}\right] \Rightarrow \angle \mathrm{ACD}=45^{\circ}\)
\(\mathrm{AD}, \angle \mathrm{BAC}=90^{\circ}\)-এর সমদ্বিখণ্ডক।
\(\therefore \angle B A D=45^{\circ}\) এবং \(\angle \mathrm{CAD}=45^{\circ}\)
এখন,
\(\frac{\sec \angle \mathrm{ACD}}{\sin \angle \mathrm{CAD}}=\frac{\sec 45^{\circ}}{\sin 45^{\circ}}\)\(\left[\because \angle \mathrm{CAD}=\frac{1}{2} \angle \mathrm{BAC}=\frac{1}{2} \times 90^{\circ}=45^{\circ}\right]\)
\(=\frac{\sqrt{2}}{\frac{1}{\sqrt{2}}}=\sqrt{2} \times \sqrt{2}=2\)
আবার, \(\operatorname{cosec}^{2} \angle \mathrm{CAD}=\operatorname{cosec}^{2} 45^{\circ}=(\sqrt{2})^{2}=2\)
\(\therefore \frac{\sec \angle \mathrm{ACD}}{\sin \angle \mathrm{CAD}}=\operatorname{cosec}^{2} \angle \mathrm{CAD}.\) (প্রমাণিত)

\(2\cos^2 \theta - 3\cos \theta + 1 = 0\)
বা, \(2\cos^2 \theta - 2\cos \theta - \cos \theta + 1 = 0\)
বা, \(2\cos \theta (\cos \theta - 1) - 1(\cos \theta - 1) = 0\)
বা, \((2\cos \theta - 1) \cdot (\cos\theta - 1) = 0\)
বা, \(2\cos\theta - 1 = 0\)
বা, \(2 \cos \theta=1\)
বা, \(\cos \theta = \frac{1}{2}\)
বা, \(\cos \theta = \cos {60^\circ }\)
বা, \(\theta=60^{\circ}\)
আবার
\(\cos \theta - 1 = 0\)
বা, \(\cos \theta = 1\)
বা, \(\cos \theta = \cos {0^\circ }\)
বা, \(\theta = {0^\circ }\)
\(\therefore\) নির্ণেয় \(\theta\)-এর মান \(\)\({0^\circ }\) ও \({60^\circ }\)