${\rm{ (i) }}\frac{{\sin {{38}^\circ }}}{{\cos {{52}^\circ }}} = \frac{{\sin \left( {{{90}^\circ } - {{52}^\circ }} \right)}}{{\cos {{52}^\circ }}} = \frac{{\cos {{52}^\circ }}}{{\cos {{52}^\circ }}} = 1$ $\left[\because \sin \left(90^{\circ}-0\right)=\cos 0\right]$
${\rm{ (ii) }}\frac{{cosec {{79}^\circ }}}{{\sec {{11}^\circ }}} = \frac{{cosec \left( {{{90}^\circ } - {{11}^\circ }} \right)}}{{\sec {{11}^\circ }}} = \frac{{\sec {{11}^\circ }}}{{\sec {{11}^\circ }}} = 1$$[\because \operatorname{cosec}(90-0)=\sec \theta]$
${\rm{ (iii) }}\frac{{\tan {{27}^\circ }}}{{\cot {{63}^\circ }}} = \frac{{\tan \left( {{{90}^\circ } - {{63}^\circ }} \right)}}{{\cot {{63}^\circ }}} = \frac{{\cot {{63}^\circ }}}{{\cot {{63}^\circ }}} = 1$$\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$

বামপক্ষ $= \sin {66^\circ } - \cos {24^\circ }$
$=\sin \left( {{{90}^\circ } - {{24}^\circ }} \right)$ $- \cos {24^\circ }$
$=\cos {24^\circ } - \cos {24^\circ }$$\left[\because \sin \left(90^{\circ}-0\right)=\cos 0\right]$
$= 0 =$ ডানপক্ষ(প্রমাণিত)

বামপক্ষ $\cos ^{2} 57^{\circ}+\cos ^{2} 35^{\circ}$
$=\cos ^{2}\left(90^{\circ}-33^{\circ}\right)+\cos ^{2} 33^{\circ}$
$=\sin ^{2} 33^{\circ}+\cos ^{2} 33^{\circ}$$\left[\because \cos ^{2}\left(90^{\circ}-\theta\right)=\sin ^{2} \theta\right]$
= 1 = ডানপক্ষ (প্রমাণিত)

বামপক্ষ = $\cos ^{2} 75^{\circ}-\sin ^{2} 15^{\circ}$
$=\cos ^{2}\left(90^{\circ}-15^{\circ}\right)-\sin ^{2} 15^\circ$ $\left[\cos ^{2}\left(90^{\circ}-\theta\right)=\sin ^{2} \theta\right]$
$=\sin ^{2} 15^{\circ}-\sin ^{2} 15=0$ = ডানপক্ষ (প্রমাণিত)

বামপক্ষ=$\operatorname{cosec}^{2} 48^{\circ}-\tan ^{2} 42^{\circ}=1$
$=\operatorname{cosec}^{2}\left(90^{\circ}-42^{\circ}\right)-\tan ^{2} 42^{\circ}$
$=\sec ^{2} 42^{\circ}-\tan ^{2} 42^{\circ}$$\left[\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)=\sec ^{2} \theta\right]$
$= 1 =$ ডানপক্ষ

$\sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}$
$=\sec 70^{\circ} \sin \left(90^{\circ}-70^{\circ}\right)+\cos 20^{\circ} \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)$
$=\sec 70^{\circ} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \sec 20^{\circ}$
$=\frac{1}{\cos 70^{\circ}} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}}$
$=1+1=2$
$\therefore \sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=2$ (প্রমাণিত)

${\sin ^2}\alpha + {\sin ^2}\beta = 1$
যদি $\alpha$ ও $\beta$ কোণদুটি পরস্পর পূরক কোণ হয়,
তাহলে দেখাই যে ${\sin ^2}\alpha + {\sin ^2}\beta = 1$
[$\because \alpha+\beta=90^{\circ}$$\quad$$\alpha$$=90^{\circ}-\beta$]
$\sin ^{2} \alpha+\sin ^{2} \beta$
$=\sin ^{2}\left(90^{\circ}-\beta\right)+\sin ^{2} \beta$
$=\cos^2\beta+\sin ^{2} \beta$$\left[\because \sin ^2\left(90^{\circ}-\theta\right)=\cos ^2 \theta\right]$
$=1$ (প্রমাণিত)

বামপক্ষ $=\cot \beta+\cos \beta$
$=\frac{\cos \beta}{\sin \beta}+\cos \beta$
$=\frac{\cos \beta}{\sin \left(90^{\circ}-\alpha\right)}+\cos \beta$
$=\frac{\cos \beta}{\cos \alpha}+\cos \beta$$\left[\because \sin \left(90^{\circ}-\theta\right) \times=\cos \theta\right]$
$=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \sin \beta}{\sin \beta}$
$=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \sin \left(90^{\circ}-\alpha\right)}{\sin \left(90^{\circ}-\alpha)\right.}$
$=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \cos \alpha}{\cos \alpha}$
$=\frac{\cos \beta}{\cos \alpha} \times (1+\cos \alpha)$
$=\frac{\cos \beta}{\cos \alpha}\left\{1+\cos \left(90^{\circ}-\beta\right)\right\}$
$=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)$
$\therefore \cot \beta+\cos \beta=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)$ (প্রমাণিত)

বামপক্ষ $\frac{{\sec \alpha }}{{\cos \alpha }} - {\cot ^2}\beta$
$= \frac{{\sec \left( {{{90}^\circ } - \beta } \right)}}{{\cos \left( {{{90}^\circ } - \beta } \right)}} - {\cot ^2}\beta$
$= \frac{{cosec \beta }}{{\sin \beta }} - {\cot ^2}\beta$$\left[\because \sec \left(90^{\circ}-13\right)=\operatorname{cosec} \beta, \cos \left(90^{\circ}-\beta\right)=\sin \beta\right]$
$= cosec^2\beta - {\cot ^2}\beta$
$= 1 =$ ডানপক্ষ (প্রমাণিত)

প্রথম পদ্ধতি :
$\therefore \cos 17^{0}=\sqrt{1-\sin ^{2} 17^{0}}=\sqrt{1-\frac{x^{2}}{y^{2}}}$
[ যেহেতু, $\sin 17^{\circ}=\frac{x}{y}$]
$=\sqrt{\frac{y^{2}-x^{2}}{y^{2}}}=\frac{\sqrt{y^{2}-x^{2}}}{y}$
$\therefore \quad \sec 17^{0}=\frac{1}{\cos 17^{\circ}}=\frac{y}{\sqrt{y^{2}-x^{2}}}$
আবার, $\sin 73^{\circ}=\sin \left(90^{\circ}-17^{\circ}\right)=\cos 17^{0}=\frac{\sqrt{y^{2}-x^{2}}}{y}$$\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$\therefore \sec 17^{0}-\sin 73^{\circ}=\frac{y}{\sqrt{y^{2}-x^{2}}}-\frac{\sqrt{y^{2}-x^{2}}}{y}$
$=\frac{y^{2}-\left(\sqrt{y^{2}-x^{2}}\right)^{2}}{y \sqrt{y^{2}-x^{2}}}=\frac{y^{2}-y^{2}+x^{2}}{y \sqrt{y^{2}-x^{2}}}=\frac{x^{2}}{y \sqrt{y^{2}-x^{2}}}$ (প্রমানিত)
দ্বিতীয় পদ্ধতি :
$\sec 17^{\circ}-\sin 73^{\circ}$
$=\frac{1}{\cos 17^{0}}-\cos 17^{\circ} \quad\left[\because \sin 73^{0}=\sin \left(90^{\circ}-17^{\circ}\right)=\cos 17^{0}\right]$
$=\frac{1-\cos ^{2} 17^{0}}{\cos 17^{0}} \quad=\frac{\sin ^{2} 17^{0}}{\sqrt{1-\sin ^{2} 17^{0}}}$
$=\frac{\left(\frac{x}{y}\right)^{2}}{\sqrt{1-\left(\frac{x}{y}\right)^{2}}} \quad\left[\because \sin 17^{0}=\frac{x}{y}\right]$
$=\frac{\frac{x^{2}}{y^{2}}}{\sqrt{\frac{y^{2}-x^{2}}{y^{2}}}}=\frac{x^{2}}{y^{2}} \cdot \frac{y}{\sqrt{y^{2}-x^{2}}}=\frac{x^{2}}{y \sqrt{y^{2}-x^{2}}}$(প্রমানিত)

$\sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}$
$=\sec ^{2} 12^{\circ}-\cot ^{2} 78^{\circ}.$
$=\sec ^{2} 12^{\circ}-\cot ^{2}\left(90^{\circ}-12^{\circ}\right)$
$=\sec ^{2} 12^{\circ}-\tan ^{2} 12^{\circ}$$\left[\because \cot \left(90^{\circ}-0\right)=\tan \theta\right]$
$=1$
$\therefore \sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}=1$ (প্রমাণিত)

সমাধান : বামপক্ষ $1 + \frac{{\tan A}}{{\tan B}}$
$= 1 + \frac{{\tan A}}{{\tan \left( {{{90}^\circ } - A} \right)}}$ $\left[ {\therefore \angle A + \angle B = {{90}^\circ }{\rm{ or }}\angle B = {{90}^\circ } - \angle A} \right]$
$= 1 + \frac{{\tan A}}{{\cot A}} = 1 + \tan A \times \tan A=1+\tan ^{2} A$
$= Sec{ ^2}A =$ ডানপক্ষ (প্রমাণিত)

বামপক্ষ $=\operatorname{cosec}^{2} 22^{\circ} \cot ^{2} 68^{\circ}$
$=\operatorname{cosec}^{2}\left(90^{\circ}-68^{\circ}\right) \cot ^{2} 68^{\circ}$$\left[ \because \operatorname{cosec}\left(90^{\circ}-\theta\right) = \sec \theta\right]$
$=\sec ^{2} 68^{\circ} \cot ^{2} 68^{\circ}=\left(1+\tan ^{2} 68^{\circ}\right) \cot ^{2} 68^{\circ}$$\left[ \because\sec ^{2} \theta=1+\tan ^{2} \theta\right]$
$=\cot ^{2} 68^{\circ}+\tan ^{2} 68^{\circ} \times \frac{1}{\tan ^{2} 68^{\circ}}$
$=\cot ^{2} 68^{\circ}+1=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}$$\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\cos ^{2}\left(90^{\circ}-68^{\circ}\right)$$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}=\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 68^{\circ}$ = ডানপক্ষ
$\therefore$ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

ডানপক্ষ$=\sqrt{\frac{\sin P}{\cos Q}-\sin P \cos Q}$
$=\sqrt{\frac{\sin P}{\cos \left(90^{\circ}-P\right)}-\sin P \cos \left(90^{\circ}-P\right)}$ $\left[P+Q=90^{\circ} \Rightarrow Q=90^{\circ}-P\right]$
$=\sqrt{\frac{\sin P}{\sin P}-\sin P \sin P}$ $\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$=\sqrt{1-\sin ^{2} \mathrm{P}}$
$=\sqrt{\cos ^{2} \mathrm{P}}$
$=\cos P=$ বামপক্ষ (প্রমানিত)

$\cot 12^{\circ} \cot 38^{\circ} \cot 52^{\circ} \cot 78^{\circ} \cot 60^{\circ}$
$=\left(\cot 12^{\circ} \cot 78^{\circ}\right)\left(\cot 38^{\circ} \cot 52^{\circ}\right) \cot 60^{\circ} .$
$=\left\{\cot 12^{\circ} \times \cot \left(90^{\circ}-12^{\circ}\right)\right\}\left\{\cot 38^{\circ} \cdot \cot \left(90^{\circ}-38^{\circ}\right)\right\} \cot 60^{\circ}$
$=\left(\cot 12^{\circ} \cdot \tan 12^{\circ}\right)\left(\cot 38^{\circ} \cdot \tan 38^{\circ}\right) \cot 60^{\circ} .$$\left[\because \cos \left(90^{\circ}-\theta\right)=\tan \theta\right]$
$=\left(\frac{1}{\tan 12^{\circ}} \times \tan 12^{\circ}\right)\left(\frac{1}{\tan 38^{\circ}} \times \tan 38^{\circ}\right) \cot 60^{\circ}$
$=1 \times 1 \times \cot 60^{\circ}=\cot 60^{\circ}$
$=\frac{1}{\sqrt{3}}$
$\therefore \cot 12^{\circ} \cot 38^{\circ} \cot 52^{\circ} \cot 78^{\circ} \cot 60^{\circ}=\frac{1}{\sqrt{3}}$ (প্রমাণিত)

সমাধান : $\angle ACB$ অর্ধবৃত্তস্থ কোণ
${\rm{ (i) }}\tan \angle {\rm{ABC}}$
$\tan({90^\circ-\angle ACO})$
$=\cot\angle ACO[\because OB=OC]$ $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$ (প্রমাণিত)
${\rm{ (ii) }}{\sin ^2}\angle {\rm{BCO}} + {\sin ^2}\angle {\rm{ACO}}$
$= {\sin ^2}\left\{ {{{90}^\circ } - \angle ACO} \right\} + {\sin ^2}\angle ACO$ $= {\cos ^2}\angle ACO + {\sin ^2}\angle ACO = 1 =$ ডানপক্ষ (প্রমাণিত)
${\rm{ (iii) }}{cosec ^2} \angle CAB - 1$
$= {cosec ^2}\left\{ {{{90}^\circ }-\angle ABC} \right\} - 1$
$= {\sec ^2}\angle ABC – 1$ $\left[\because \sec ^{2} \theta-1=\tan ^{2} \theta\right]$
$= {\tan ^2}\angle ABC =$ ডানপক্ষ (প্রমাণিত)

সমাধান : $\tan \angle ACD = \tan \left( {{{90}^\circ } - \angle ACB} \right)$
$= \cot \angle ACB$ (প্রমাণিত)

${\tan ^2} \angle CAD + 1 = {\tan ^2}\left( {{{90}^\circ } - \angle BAC} \right) + 1$$\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$
$= {\cot ^2}\angle BAC + 1 = {cosec ^2}\angle BAC$ (প্রমাণিত) $\left[\because \cot ^{2} \theta+1=\operatorname{cosec}^{2} \theta\right]$

সমাধান : $\sin {43^\circ }\cos {47^\circ } + \cos {43^\circ } \cdot \sin {47^\circ }$
$=\sin \left(90^{\circ}-47^{\circ}\right) \cdot \cos 47+\cos \left(90^{\circ}-47^{\circ}\right) \cdot \sin 47$
$=\cos 47 \times \cos 47+\sin 47 \times \sin 47^{\circ}$
$=\cos ^{2} 47+\sin ^{2} 47^{\circ}=1$
(b) 1

$\left( {\frac{{\tan {{35}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\cot {{78}^\circ }}}{{\tan {{12}^\circ }}}} \right)$
$= \left( {\frac{{\tan \left( {{{90}^\circ } - {{55}^\circ }} \right)}}{{\cot {{55}^\circ }}} + \frac{{\cot \left( {{{90}^\circ } - {{12}^\circ }} \right)}}{{ \tan {{12}^\circ }}}} \right)$
$= \frac{{\cot {{55}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\tan {{12}^\circ }}}{{\tan {{12}^\circ }}}$$\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta, \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]$
$= 1 + 1 = 2$
(c) 2

$\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)$
$=\cos \left(40^{\circ}+\theta\right)-\sin \left\{90^{\circ}-\left(40^{\circ}+\theta\right)\right\}$
$=\cos \left(40^{\circ}+\theta\right)-\cos \left(40^{\circ}+\theta\right)$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$=0 .$
$\therefore$ (c) উত্তরটি সঠিক।

$\because \mathrm{ABC}$ একটি ত্রিভুজ,
$\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
বা, $\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}-\angle \mathrm{A}$
বা, $\frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}=\frac{180^{\circ}}{2}-\frac{\angle \mathrm{A}}{2}$
$\Rightarrow \frac{\mathrm{B}+\mathrm{C}}{2}=90^{\circ}-\frac{\mathrm{A}}{2}$
$\therefore \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)$
$=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)$
$=\cos \frac{\mathrm{A}}{2}$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
$\therefore$ (b) উত্তরটি সঠিক।

$\cot B=\cot(90^\circ -A)=\tan A=\frac{3}{4}$
${\rm{ (a) }}\frac{3}{4}$

$\cos {54^\circ } = \cos \left( {90 - {{36}^\circ }} \right) = \sin {36^\circ }$$\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]$
$\therefore \cos {54^\circ } = \sin {36^\circ }$
সত্য

$\sin {12^\circ } - \cos {78^\circ }$
$= \sin \left( {{{90}^\circ } - {{78}^\circ }} \right) - \cos {78^\circ }$
$= \cos {{78}^\circ } - \cos {{78}^\circ }= 0$$\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
মিথ্যা

$\tan {15^\circ } \times \tan {45^\circ } \times \tan {60^\circ }\tan \times {75^\circ }$
$= \tan \left( {{{90}^\circ } - 75} \right) \times 1 \times \sqrt 3 \times \tan {75^\circ }$
$=\cot 75^{\circ} \times 1 \times \sqrt{3} \times \tan 75^{\circ}\left[\tan \left(90^{\circ}-0\right)=\cot \theta\right]$
$=\sqrt{3} \times \frac{1}{\tan 75^{\circ}} \times \tan 75^{\circ}$
$= \sqrt 3$
Ans. $\sqrt 3$

$\sin 12^{\circ} \cos 18^{\circ} \sec 78^{\circ} \operatorname{cosec} 72^{\circ}$
$\Rightarrow \sin \left(90^{\circ}-78^{\circ}\right) \cos \left(90^{\circ}-72^{\circ}\right) \operatorname{sec} 78^{\circ} \operatorname{cosec} 72^{\circ}$
$\quad\quad\left[\because \sin \left(90^{\circ}-\theta\right) =\cos \theta ও \cos \left(90^{\circ}-\theta\right) =\sin \theta\right]$
$=\cos 78^{\circ}\cdot \sin 72^{\circ} \cdot \operatorname{sec} 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}$
$\Rightarrow \frac{1}{\sec 78^{\circ}} \cdot \frac{1}{\operatorname{cosec} 72^{\circ}} \cdot \sec 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}$
$=1$
$\therefore$ নির্ণেয় মান $=1$

সমাধান : $A + B = {90^\circ }$
$\therefore A = \left( {{{90}^\circ } - B} \right)$
$\sin A = \sin \left( {{{90}^\circ } - B} \right)$
$= \cos B$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
Ans. $\cos B$

$\sin 10\theta = \cos 8\theta$
বা, $\sin 10\theta = \sin \left( {{{90}^\circ } - 8\theta } \right)$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
বা, $10\theta = {90^\circ } - 8\theta$
বা, $100+8 \theta=90^{\circ}$
বা, $18\theta = {90^\circ }$
বা, $\theta=\frac{90^{\circ}}{5}$
বা, $\theta = {5^\circ }$
$\tan 9\theta = \tan \left(9 \times 5^{\circ}\right)$
$= \tan {45^\circ } = 1$
$\therefore$ নির্ণেয় $10\theta$ এর মান = 1

$\tan 4\theta \times \tan 6\theta = 1$
বা, $\tan 4\theta = \frac{1}{{\tan 6\theta }}$
বা, $\tan 4\theta = \cot 6\theta$
বা, $\tan 4\theta = \tan \left( {{{90}^\circ } - 6\theta } \right)$ $\left[\because \tan \left(90^{\circ}-0\right)=\cot 0\right]$
বা, $4\theta + 6\theta = {90^\circ }$
বা, $100=90^{\circ}$
বা, $0=\frac{90^{\circ}}{10}$
বা, $\theta = {9^\circ }$
$\therefore \boldsymbol{\theta}$ মান হল $9^{\circ}$

$\frac{2 \sin ^{2} 63^{\circ}+1+2 \sin ^{2} 27^{\circ}}{3 \cos ^{2} 17^{\circ}-2+3 \cos ^{2} 73^{\circ}}$
$=\frac{2 \sin ^{2} 63^{\circ}+2 \sin ^{2}\left(90^{\circ}-63^{\circ}\right)+1}{3 \cos ^{2} 17^{\circ}+3 \cos ^{2}\left(90^{\circ}-17^{\circ}\right)-2}$
$=\frac{2 \sin ^{2} 63^{\circ}+2 \cos ^{2} 63^{\circ}+1}{3 \cos ^{2} 17^{\circ}+3 \sin ^{2} 17^{\circ}-2}$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta, \cos \left(90^{\circ}-\theta\right)=\sin 0\right]$
$=\frac{2\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)+1}{3\left(\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}\right)-2}$
$=\frac{2 \times 1+1}{3 \times 1-2}=\frac{2+1}{3-2}=\frac{3}{1}=3 .$
$\therefore$ প্রদত্ত রাশির মান $= 3$.

$\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 89^{\circ}$
$=\tan 1^{\circ}+\tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ}$
$\quad\quad\times \ldots \times \tan 87^{\circ} \times \tan 88^{\circ} \times \tan 89^{\circ}$

$=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ} \times \ldots$
$\quad\quad\times \tan \left(90^{\circ}-3^{\circ}\right) \times \tan \left(90^{\circ}-2^{\circ}\right) \times \tan \left(90^{\circ}-1^{\circ}\right)$

$=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times\tan 45^{\circ} \times \cot 3^{\circ}$
$\quad\quad\times \cot 2^{\circ} \times \cot 1^{\circ} \quad$[যেহেতু $\tan \left(90^{\circ}-\alpha\right)=\cot \alpha$]

$=\left(\tan 1^{\circ} \times \cot 1^{\circ}\right) \times\left(\tan 2^{\circ} \times \cot 2^ \circ\right)$
$\quad\quad\times\left(\tan 3^{\circ} \times \cot 3^{\circ}\right) \times \ldots \times \tan 45^{\circ}$

$=1 \times 1 \times 1 \times \cdots \times \tan 45^{\circ}$
$\quad\quad [$যেহেতু $\tan \theta \times \cot \theta=1]$
$=\tan 45^{\circ}$
$= 1$

দেওয়া আছে, $\sec 5 A=\operatorname{cosec}\left(A+36^{\circ}\right)$
বা, $\operatorname{cosec}\left(90^{\circ}-5 A\right)=\operatorname{cosec}\left(A+36^{\circ}\right)$
$\Rightarrow 90^{\circ}-5 \mathrm{~A}=\mathrm{A}+36^{\circ}$
$\Rightarrow 5 \mathrm{~A}+\mathrm{A}=90^{\circ}-36^{\circ}$
বা, $6 \mathrm{~A}=54^{\circ}$
বা, $\mathrm{A}=\frac{54^{\circ}}{6}$
বা, $A=9^{\circ}$
$\therefore$ $A$-এর নির্ণেয় মান $= 9^{\circ}$.