WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | Koshe Dekhi 24 Class10 | Trigonometric Ratios Of Complementary Angle | WBBSE Madhyamik Ganit Prakash Class 10(Ten) (X) Math Solution Of Chapter 24 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি ২৪
Share this page using :
WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি
২৪
কষে দেখি - 24
WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি ২৪
1. মান নির্ণয় করি : \(({\rm{i}}){\rm{ }}\frac{{\sin {{38}^\circ }}}{{\cos {{52}^\circ }}}{\rm{ }}({\rm{ii}}){\rm{ }}\frac{{cosec {{79}^\circ }}}{{\sec {{11}^\circ }}}{\rm{ }}({\rm{iii}}){\rm{ }}\frac{{\tan {{27}^\circ }}}{{\cot {{63}^\circ }}}\)
\({\rm{ (i) }}\frac{{\sin {{38}^\circ }}}{{\cos {{52}^\circ }}} = \frac{{\sin \left( {{{90}^\circ } - {{52}^\circ }} \right)}}{{\cos {{52}^\circ }}} = \frac{{\cos {{52}^\circ }}}{{\cos {{52}^\circ }}} = 1\) \(\left[\because \sin \left(90^{\circ}-0\right)=\cos 0\right]\)
\({\rm{ (ii) }}\frac{{cosec {{79}^\circ }}}{{\sec {{11}^\circ }}} = \frac{{cosec \left( {{{90}^\circ } - {{11}^\circ }} \right)}}{{\sec {{11}^\circ }}} = \frac{{\sec {{11}^\circ }}}{{\sec {{11}^\circ }}} = 1\)\([\because \operatorname{cosec}(90-0)=\sec \theta]\)
\({\rm{ (iii) }}\frac{{\tan {{27}^\circ }}}{{\cot {{63}^\circ }}} = \frac{{\tan \left( {{{90}^\circ } - {{63}^\circ }} \right)}}{{\cot {{63}^\circ }}} = \frac{{\cot {{63}^\circ }}}{{\cot {{63}^\circ }}} = 1\)\(\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]\)
\({\rm{ (ii) }}\frac{{cosec {{79}^\circ }}}{{\sec {{11}^\circ }}} = \frac{{cosec \left( {{{90}^\circ } - {{11}^\circ }} \right)}}{{\sec {{11}^\circ }}} = \frac{{\sec {{11}^\circ }}}{{\sec {{11}^\circ }}} = 1\)\([\because \operatorname{cosec}(90-0)=\sec \theta]\)
\({\rm{ (iii) }}\frac{{\tan {{27}^\circ }}}{{\cot {{63}^\circ }}} = \frac{{\tan \left( {{{90}^\circ } - {{63}^\circ }} \right)}}{{\cot {{63}^\circ }}} = \frac{{\cot {{63}^\circ }}}{{\cot {{63}^\circ }}} = 1\)\(\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]\)
2. দেখাই যে :
(i) \( \sin {{66}^\circ } - \cos {{24}^\circ } = 0\)
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)
\(=\sin \left( {{{90}^\circ } - {{24}^\circ }} \right)\) \( - \cos {24^\circ }\)
\(=\cos {24^\circ } - \cos {24^\circ }\)\(\left[\because \sin \left(90^{\circ}-0\right)=\cos 0\right]\)
\( = 0 = \) ডানপক্ষ(প্রমাণিত)
(ii) \( {{\cos }^2}{{57}^\circ } - {{\cos }^2}{{33}^\circ } = 1\)
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)
\(=\cos ^{2}\left(90^{\circ}-33^{\circ}\right)+\cos ^{2} 33^{\circ}\)
\(=\sin ^{2} 33^{\circ}+\cos ^{2} 33^{\circ}\)\(\left[\because \cos ^{2}\left(90^{\circ}-\theta\right)=\sin ^{2} \theta\right]\)
= 1 = ডানপক্ষ (প্রমাণিত)
(iii) \(\cos ^{2} 75^{\circ}-\sin ^{2} 15^{\circ}=0\)
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)
\(=\cos ^{2}\left(90^{\circ}-15^{\circ}\right)-\sin ^{2} 15^\circ\) \(\left[\cos ^{2}\left(90^{\circ}-\theta\right)=\sin ^{2} \theta\right]\)
\(=\sin ^{2} 15^{\circ}-\sin ^{2} 15=0\) = ডানপক্ষ (প্রমাণিত)
(iv) \(\operatorname{cosec}^{2} 48^{\circ}-\tan ^{2} 42^{\circ}=1\)
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
\(=\operatorname{cosec}^{2}\left(90^{\circ}-42^{\circ}\right)-\tan ^{2} 42^{\circ}\)
\(=\sec ^{2} 42^{\circ}-\tan ^{2} 42^{\circ}\)\(\left[\operatorname{cosec}^{2}\left(90^{\circ}-\theta\right)=\sec ^{2} \theta\right]\)
\( = 1 =\) ডানপক্ষ
(v) \(\sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=2\)
\(\sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}\)
\(=\sec 70^{\circ} \sin \left(90^{\circ}-70^{\circ}\right)+\cos 20^{\circ} \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)\)
\(=\sec 70^{\circ} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \sec 20^{\circ}\)
\(=\frac{1}{\cos 70^{\circ}} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}}\)
\(=1+1=2\)
\(\therefore \sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=2\) (প্রমাণিত)
\(=\sec 70^{\circ} \sin \left(90^{\circ}-70^{\circ}\right)+\cos 20^{\circ} \operatorname{cosec}\left(90^{\circ}-20^{\circ}\right)\)
\(=\sec 70^{\circ} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \sec 20^{\circ}\)
\(=\frac{1}{\cos 70^{\circ}} \cdot \cos 70^{\circ}+\cos 20^{\circ} \cdot \frac{1}{\cos 20^{\circ}}\)
\(=1+1=2\)
\(\therefore \sec 70^{\circ} \sin 20^{\circ}+\cos 20^{\circ} \operatorname{cosec} 70^{\circ}=2\) (প্রমাণিত)
3. যদি \(\alpha\) ও \(\beta\) কোণদুটি পরস্পর পূরক কোণ হয়, তাহলে দেখাই যে,
(i) \( {\sin ^2}\alpha + {\sin ^2}\beta = 1\)
\({\sin ^2}\alpha + {\sin ^2}\beta = 1\)
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)
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)
(ii) \(\cot \beta+\cos \beta=\frac{\cos \beta}{\cos \alpha}\)\((1+\sin \beta)\)
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)
\(=\frac{\cos \beta}{\sin \beta}+\cos \beta\)
\(=\frac{\cos \beta}{\sin \left(90^{\circ}-\alpha\right)}+\cos \beta\)
\(=\frac{\cos \beta}{\cos \alpha}+\cos \beta\)\(\left[\because \sin \left(90^{\circ}-\theta\right) \times=\cos \theta\right]\)
\(=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \sin \beta}{\sin \beta}\)
\(=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \sin \left(90^{\circ}-\alpha\right)}{\sin \left(90^{\circ}-\alpha)\right.}\)
\(=\frac{\cos \beta}{\cos \alpha}+\frac{\cos \beta \times \cos \alpha}{\cos \alpha}\)
\(=\frac{\cos \beta}{\cos \alpha} \times (1+\cos \alpha)\)
\(=\frac{\cos \beta}{\cos \alpha}\left\{1+\cos \left(90^{\circ}-\beta\right)\right\}\)
\(=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\)
\(\therefore \cot \beta+\cos \beta=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\) (প্রমাণিত)
(iii) \(\frac{{\sec \alpha }}{{\cos \alpha }} - {\cot ^2}\beta = 1\)
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)
\(= \frac{{\sec \left( {{{90}^\circ } - \beta } \right)}}{{\cos \left( {{{90}^\circ } - \beta } \right)}} - {\cot ^2}\beta\)
\( = \frac{{cosec \beta }}{{\sin \beta }} - {\cot ^2}\beta \)\(\left[\because \sec \left(90^{\circ}-13\right)=\operatorname{cosec} \beta, \cos \left(90^{\circ}-\beta\right)=\sin \beta\right]\)
\( = cosec^2\beta - {\cot ^2}\beta\)
\( = 1 = \) ডানপক্ষ (প্রমাণিত)
4. যদি \(\sin {17^\circ } = \frac{x}{y}\) হয়, তাহলে দেখাই যে, \(\sec {17^\circ } - \sin {73^\circ } = \frac{{{{ {{x^2}}}}}}{{y\sqrt {{y^2} - {x^2}} }}\)
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)
\(\therefore \cos 17^{0}=\sqrt{1-\sin ^{2} 17^{0}}=\sqrt{1-\frac{x^{2}}{y^{2}}}\)
[ 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)
5. দেখাই যে, \({\sec ^2}{12^\circ } - \frac{1}{{{{\tan }^2}{{78}^\circ }}} = 1\)
\(\sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}\)
\(=\sec ^{2} 12^{\circ}-\cot ^{2} 78^{\circ}.\)
\(=\sec ^{2} 12^{\circ}-\cot ^{2}\left(90^{\circ}-12^{\circ}\right)\)
\(=\sec ^{2} 12^{\circ}-\tan ^{2} 12^{\circ}\)\(\left[\because \cot \left(90^{\circ}-0\right)=\tan \theta\right]\)
\(=1\)
\(\therefore \sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}=1\) (প্রমাণিত)
\(=\sec ^{2} 12^{\circ}-\cot ^{2} 78^{\circ}.\)
\(=\sec ^{2} 12^{\circ}-\cot ^{2}\left(90^{\circ}-12^{\circ}\right)\)
\(=\sec ^{2} 12^{\circ}-\tan ^{2} 12^{\circ}\)\(\left[\because \cot \left(90^{\circ}-0\right)=\tan \theta\right]\)
\(=1\)
\(\therefore \sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}=1\) (প্রমাণিত)
6. \(\angle A + \angle B = {90^\circ }\) হলে দেখাই যে \(1 + \frac{{\tan A}}{{\tan B}} = Sec{ ^2}A\)
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)
\( = 1 + \frac{{\tan A}}{{\tan \left( {{{90}^\circ } - A} \right)}}\) \(\left[ {\therefore \angle A + \angle B = {{90}^\circ }{\rm{ or }}\angle B = {{90}^\circ } - \angle A} \right]\)
\( = 1 + \frac{{\tan A}}{{\cot A}} = 1 + \tan A \times \tan A=1+\tan ^{2} A\)
\( = Sec{ ^2}A = \) ডানপক্ষ (প্রমাণিত)
WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | Koshe Dekhi 24 Class10 | Trigonometric Ratios Of Complementary Angle | WBBSE Madhyamik Ganit Prakash Class 10(Ten) (X) Math Solution Of Chapter 24 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি ২৪
7. দেখাই যে, \({cosec ^2}{22^\circ }{\cot ^2}68^{\circ} = {\sin ^2}{22^\circ } + {\sin ^2}{68^\circ } + {\cot ^2}{68^\circ }\)
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)
\(=\operatorname{cosec}^{2}\left(90^{\circ}-68^{\circ}\right) \cot ^{2} 68^{\circ}\)\(\left[ \because \operatorname{cosec}\left(90^{\circ}-\theta\right) = \sec \theta\right]\)
\(=\sec ^{2} 68^{\circ} \cot ^{2} 68^{\circ}=\left(1+\tan ^{2} 68^{\circ}\right) \cot ^{2} 68^{\circ}\)\(\left[ \because\sec ^{2} \theta=1+\tan ^{2} \theta\right]\)
\(=\cot ^{2} 68^{\circ}+\tan ^{2} 68^{\circ} \times \frac{1}{\tan ^{2} 68^{\circ}}\)
\(=\cot ^{2} 68^{\circ}+1=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}\)\(\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]\)
\(=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\cos ^{2}\left(90^{\circ}-68^{\circ}\right)\)\(\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]\)
\(=\cot ^{2} 68^{\circ}+\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}=\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}+\cot ^{2} 68^{\circ}\) = ডানপক্ষ
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)
8. যদি \(\angle P + \angle Q = {90^\circ }\) হয়, তবে দেখাই যে, \(\sqrt {\frac{{\sin P}}{{\cos Q}} - \sin P\cos Q = }\cos P \)
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)
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)
9. প্রমাণ করি যে \(\cot {12^\circ }\cot {38^\circ } \cdot \cot {52^\circ }\cot {78^\circ }\cot {60^\circ } = \frac{1}{{\sqrt 3 }}\)
\(\cot 12^{\circ} \cot 38^{\circ} \cot 52^{\circ} \cot 78^{\circ} \cot 60^{\circ}\)
\(=\left(\cot 12^{\circ} \cot 78^{\circ}\right)\left(\cot 38^{\circ} \cot 52^{\circ}\right) \cot 60^{\circ} .\)
\(=\left\{\cot 12^{\circ} \times \cot \left(90^{\circ}-12^{\circ}\right)\right\}\left\{\cot 38^{\circ} \cdot \cot \left(90^{\circ}-38^{\circ}\right)\right\} \cot 60^{\circ}\)
\(=\left(\cot 12^{\circ} \cdot \tan 12^{\circ}\right)\left(\cot 38^{\circ} \cdot \tan 38^{\circ}\right) \cot 60^{\circ} .\)\(\left[\because \cos \left(90^{\circ}-\theta\right)=\tan \theta\right]\)
\(=\left(\frac{1}{\tan 12^{\circ}} \times \tan 12^{\circ}\right)\left(\frac{1}{\tan 38^{\circ}} \times \tan 38^{\circ}\right) \cot 60^{\circ}\)
\(=1 \times 1 \times \cot 60^{\circ}=\cot 60^{\circ}\)
\(=\frac{1}{\sqrt{3}}\)
\(\therefore \cot 12^{\circ} \cot 38^{\circ} \cot 52^{\circ} \cot 78^{\circ} \cot 60^{\circ}=\frac{1}{\sqrt{3}}\) (প্রমাণিত)
\(=\left(\cot 12^{\circ} \cot 78^{\circ}\right)\left(\cot 38^{\circ} \cot 52^{\circ}\right) \cot 60^{\circ} .\)
\(=\left\{\cot 12^{\circ} \times \cot \left(90^{\circ}-12^{\circ}\right)\right\}\left\{\cot 38^{\circ} \cdot \cot \left(90^{\circ}-38^{\circ}\right)\right\} \cot 60^{\circ}\)
\(=\left(\cot 12^{\circ} \cdot \tan 12^{\circ}\right)\left(\cot 38^{\circ} \cdot \tan 38^{\circ}\right) \cot 60^{\circ} .\)\(\left[\because \cos \left(90^{\circ}-\theta\right)=\tan \theta\right]\)
\(=\left(\frac{1}{\tan 12^{\circ}} \times \tan 12^{\circ}\right)\left(\frac{1}{\tan 38^{\circ}} \times \tan 38^{\circ}\right) \cot 60^{\circ}\)
\(=1 \times 1 \times \cot 60^{\circ}=\cot 60^{\circ}\)
\(=\frac{1}{\sqrt{3}}\)
\(\therefore \cot 12^{\circ} \cot 38^{\circ} \cot 52^{\circ} \cot 78^{\circ} \cot 60^{\circ}=\frac{1}{\sqrt{3}}\) (প্রমাণিত)
10. O কেন্দ্রীয় যে-কোনো একটি বৃত্তের AOB একটি ব্যাস এবং বৃত্তের উপর C যে-কোনো একটি বিন্দু, এবার A, C; B, C এবং O, C যুক্ত করে দেখাই যে,
\({\rm{ (i) }}\tan \angle ABC = {\rm{ cot }}\angle ACO\)
\({\rm{ (ii) }} {\sin ^2}\angle BCO + {\sin ^2}\) \(\angle ACO = 1\)
\({\rm{ (iii) cosec^{2} }}\angle CAB - 1 = {\tan ^2}\angle ABC\)
\({\rm{ (i) }}\tan \angle ABC = {\rm{ cot }}\angle ACO\)
\({\rm{ (ii) }} {\sin ^2}\angle BCO + {\sin ^2}\) \(\angle ACO = 1\)
\({\rm{ (iii) cosec^{2} }}\angle CAB - 1 = {\tan ^2}\angle ABC\)
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)
11. ABCD একটি আয়তাকার চিত্র। A, C যুক্ত করে প্রমাণ করি যে,
(i) \( \tan \angle ACD = \cot \angle ACB\)
সমাধান : \( \tan \angle ACD = \tan \left( {{{90}^\circ } - \angle ACB} \right)\)
\( = \cot \angle ACB\) (প্রমাণিত)
\( = \cot \angle ACB\) (প্রমাণিত)
(ii) \( {\tan ^2}\angle CAD + 1 = \frac{1}{{{{\sin }^2}\angle BAC}}\)
\({\tan ^2} \angle CAD + 1 = {\tan ^2}\left( {{{90}^\circ } - \angle BAC} \right) + 1\)\(\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]\)
\( = {\cot ^2}\angle BAC + 1 = {cosec ^2}\angle BAC\) (প্রমাণিত) \(\left[\because \cot ^{2} \theta+1=\operatorname{cosec}^{2} \theta\right]\)
\( = {\cot ^2}\angle BAC + 1 = {cosec ^2}\angle BAC\) (প্রমাণিত) \(\left[\because \cot ^{2} \theta+1=\operatorname{cosec}^{2} \theta\right]\)
12. অতিসংক্ষিপ্ত উত্তরধর্মী প্রশ্ন :
(A) বহু বিকল্পীয় প্রশ্ন (M.C.Q) :
\(({\rm{i}}){\rm{ }}\left( {\sin {{43}^\circ } \cdot \cos {{47}^\circ } + \cos {{43}^\circ }.\sin {{47}^\circ }} \right)\)-এর মান
\({\rm{ (a) }}0{\rm{ (b) }}1{\rm{ (c) }}\sin {4^\circ }{\rm{ (d) }}\cos {4^\circ }\)
\({\rm{ (a) }}0{\rm{ (b) }}1{\rm{ (c) }}\sin {4^\circ }{\rm{ (d) }}\cos {4^\circ }\)
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
\(=\sin \left(90^{\circ}-47^{\circ}\right) \cdot \cos 47+\cos \left(90^{\circ}-47^{\circ}\right) \cdot \sin 47\)
\(=\cos 47 \times \cos 47+\sin 47 \times \sin 47^{\circ}\)
\(=\cos ^{2} 47+\sin ^{2} 47^{\circ}=1\)
(b) 1
\({\rm{ (ii) }}\left( {\frac{{\tan {{35}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\cot {{78}^\circ }}}{{\tan {{12}^\circ }}}} \right)\)-এর মান
\({\rm{ (a) }}0{\rm{ (b) }}1{\rm{ (c) }}2{\rm{ (d) }}\) কোনোটিই নয়।
\({\rm{ (a) }}0{\rm{ (b) }}1{\rm{ (c) }}2{\rm{ (d) }}\) কোনোটিই নয়।
\(\left( {\frac{{\tan {{35}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\cot {{78}^\circ }}}{{\tan {{12}^\circ }}}} \right)\)
\(= \left( {\frac{{\tan \left( {{{90}^\circ } - {{55}^\circ }} \right)}}{{\cot {{55}^\circ }}} + \frac{{\cot \left( {{{90}^\circ } - {{12}^\circ }} \right)}}{{ \tan {{12}^\circ }}}} \right)\)
\( = \frac{{\cot {{55}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\tan {{12}^\circ }}}{{\tan {{12}^\circ }}}\)\(\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta, \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]\)
\( = 1 + 1 = 2\)
(c) 2
\(= \left( {\frac{{\tan \left( {{{90}^\circ } - {{55}^\circ }} \right)}}{{\cot {{55}^\circ }}} + \frac{{\cot \left( {{{90}^\circ } - {{12}^\circ }} \right)}}{{ \tan {{12}^\circ }}}} \right)\)
\( = \frac{{\cot {{55}^\circ }}}{{\cot {{55}^\circ }}} + \frac{{\tan {{12}^\circ }}}{{\tan {{12}^\circ }}}\)\(\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta, \cot \left(90^{\circ}-\theta\right)=\tan \theta\right]\)
\( = 1 + 1 = 2\)
(c) 2
\({\rm{ (iii) }}\left\{ {\cos \left( {{{40}^\circ } + \theta } \right) - \sin \left( {{{50}^\circ } - \theta } \right)} \right\}\)-এর মান
\({\rm{ (a) }}2\cos \theta {\rm{ (b) }}7\sin \theta {\rm{ (c) }}0{\rm{ (d) }}1\)
\({\rm{ (a) }}2\cos \theta {\rm{ (b) }}7\sin \theta {\rm{ (c) }}0{\rm{ (d) }}1\)
\(\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)\)
\(=\cos \left(40^{\circ}+\theta\right)-\sin \left\{90^{\circ}-\left(40^{\circ}+\theta\right)\right\}\)
\(=\cos \left(40^{\circ}+\theta\right)-\cos \left(40^{\circ}+\theta\right)\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
\(=0 .\)
\(\therefore\) (c) উত্তরটি সঠিক।
\(=\cos \left(40^{\circ}+\theta\right)-\sin \left\{90^{\circ}-\left(40^{\circ}+\theta\right)\right\}\)
\(=\cos \left(40^{\circ}+\theta\right)-\cos \left(40^{\circ}+\theta\right)\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
\(=0 .\)
\(\therefore\) (c) উত্তরটি সঠিক।
\((iv) ABC \) একটি ত্রিভুজ । \(\sin \left( {\frac{{B + C}}{2}} \right)\) =
\((a)\sin \frac{A}{2}(b)\cos \frac{A}{2}(c)\sin A(d)\cos A\)
\((a)\sin \frac{A}{2}(b)\cos \frac{A}{2}(c)\sin A(d)\cos A\)
\(\because \mathrm{ABC}\) 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।
\(\therefore \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}\)
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।
\({\rm{ (v) }}A + B = {90^\circ }\) এবং \(\tan A = \frac{3}{4}\) হলে, \(\cot B \)-এর মান
\({\rm{ (a) }}\frac{3}{4}{\rm{ (b) }}\frac{4}{3}{\rm{ (c) }}\frac{3}{5}{\rm{ (d) }}\frac{4}{5}\)
\({\rm{ (a) }}\frac{3}{4}{\rm{ (b) }}\frac{4}{3}{\rm{ (c) }}\frac{3}{5}{\rm{ (d) }}\frac{4}{5}\)
\(\cot B=\cot(90^\circ -A)=\tan A=\frac{3}{4}\)
\({\rm{ (a) }}\frac{3}{4}\)
\({\rm{ (a) }}\frac{3}{4}\)
(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি :
\({\rm{ (i) }} \cos {54^\circ }\) এবং \(\sin {36^\circ }\)-এর মান সমান।
\(\cos {54^\circ } = \cos \left( {90 - {{36}^\circ }} \right) = \sin {36^\circ }\)\(\left[\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right]\)
\(\therefore \cos {54^\circ } = \sin {36^\circ }\)
সত্য
\(\therefore \cos {54^\circ } = \sin {36^\circ }\)
সত্য
\({\rm{ (ii) }} \left( {\sin {{12}^\circ } - \cos {{78}^\circ }} \right)\) -এর সরলতম মান 1
\(\sin {12^\circ } - \cos {78^\circ }\)
\( = \sin \left( {{{90}^\circ } - {{78}^\circ }} \right) - \cos {78^\circ }\)
\(= \cos {{78}^\circ } - \cos {{78}^\circ }= 0\)\(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
মিথ্যা
\( = \sin \left( {{{90}^\circ } - {{78}^\circ }} \right) - \cos {78^\circ }\)
\(= \cos {{78}^\circ } - \cos {{78}^\circ }= 0\)\(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
মিথ্যা
(C) শূন্যস্থান পূরণ করি :
\({\rm{(i) }}(\tan {15^\circ } \times \tan {45^\circ } \times \tan {60^\circ } \times \tan {75^\circ })\)-এর মান ________।
\(\tan {15^\circ } \times \tan {45^\circ } \times \tan {60^\circ }\tan \times {75^\circ }\)
\( = \tan \left( {{{90}^\circ } - 75} \right) \times 1 \times \sqrt 3 \times \tan {75^\circ }\)
\(=\cot 75^{\circ} \times 1 \times \sqrt{3} \times \tan 75^{\circ}\left[\tan \left(90^{\circ}-0\right)=\cot \theta\right]\)
\(=\sqrt{3} \times \frac{1}{\tan 75^{\circ}} \times \tan 75^{\circ}\)
\( = \sqrt 3 \)
Ans. \(\sqrt 3\)
\( = \tan \left( {{{90}^\circ } - 75} \right) \times 1 \times \sqrt 3 \times \tan {75^\circ }\)
\(=\cot 75^{\circ} \times 1 \times \sqrt{3} \times \tan 75^{\circ}\left[\tan \left(90^{\circ}-0\right)=\cot \theta\right]\)
\(=\sqrt{3} \times \frac{1}{\tan 75^{\circ}} \times \tan 75^{\circ}\)
\( = \sqrt 3 \)
Ans. \(\sqrt 3\)
\({\rm{ (ii) }}\left( {\sin {{12}^\circ } \times \cos {{18}^\circ }\times \sec {{78}^\circ } \times cosec {{72}^\circ }} \right)\)-এর মান ________।
\(\sin 12^{\circ} \cos 18^{\circ} \sec 78^{\circ} \operatorname{cosec} 72^{\circ}\)
\(\Rightarrow \sin \left(90^{\circ}-78^{\circ}\right) \cos \left(90^{\circ}-72^{\circ}\right) \operatorname{sec} 78^{\circ} \operatorname{cosec} 72^{\circ}\)
\(\quad\quad\left[\because \sin \left(90^{\circ}-\theta\right) =\cos \theta ও \cos \left(90^{\circ}-\theta\right) =\sin \theta\right]\)
\(=\cos 78^{\circ}\cdot \sin 72^{\circ} \cdot \operatorname{sec} 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}\)
\(\Rightarrow \frac{1}{\sec 78^{\circ}} \cdot \frac{1}{\operatorname{cosec} 72^{\circ}} \cdot \sec 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}\)
\(=1\)
\(\therefore\) নির্ণেয় মান \(=1\)
\(\Rightarrow \sin \left(90^{\circ}-78^{\circ}\right) \cos \left(90^{\circ}-72^{\circ}\right) \operatorname{sec} 78^{\circ} \operatorname{cosec} 72^{\circ}\)
\(\quad\quad\left[\because \sin \left(90^{\circ}-\theta\right) =\cos \theta ও \cos \left(90^{\circ}-\theta\right) =\sin \theta\right]\)
\(=\cos 78^{\circ}\cdot \sin 72^{\circ} \cdot \operatorname{sec} 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}\)
\(\Rightarrow \frac{1}{\sec 78^{\circ}} \cdot \frac{1}{\operatorname{cosec} 72^{\circ}} \cdot \sec 78^{\circ} \cdot \operatorname{cosec} 72^{\circ}\)
\(=1\)
\(\therefore\) নির্ণেয় মান \(=1\)
WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | Koshe Dekhi 24 Class10 | Trigonometric Ratios Of Complementary Angle | WBBSE Madhyamik Ganit Prakash Class 10(Ten) (X) Math Solution Of Chapter 24 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি ২৪
\((iii)\) A এবং B পরস্পর পূরক কোণ হলে, \(\sin A\) = _______।
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. \(\cos B\)
\(\therefore A = \left( {{{90}^\circ } - B} \right)\)
\( \sin A = \sin \left( {{{90}^\circ } - B} \right)\)
\( = \cos B\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\)
Ans. \(\cos B\)
(13) সংক্ষিপ্ত উত্তরধর্মী প্ৰশ্ন (S.A) :
\({\rm{ (i) }}\sin 10\theta = \cos 8\theta \) এবং \(10\theta \) ধনাত্মক সূক্ষ্মকোণ হলে, \(\tan 9\theta\) এর মান নির্ণয় করি।
\(\sin 10\theta = \cos 8\theta \)
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
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
\({\rm{ (ii) }}\tan 4\theta \times \tan 6\theta = 1\) এবং \( 6\theta\) ধনাত্মক সূক্ষ্মকোণ হলে, \(\theta\)-এর মান নির্ণয় করি ।
\(\tan 4\theta \times \tan 6\theta = 1\)
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 \(9^{\circ}\)
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 \(9^{\circ}\)
\({\rm{ (iii) }}\frac{{2{{\sin }^2}63 + 1 + 2{{\sin }^2}{{27}^\circ }}}{{3{{\cos }^2}{{17}^\circ } - 2 + 3{{\cos }^2}{{73}^\circ }}}\) এর মান নির্ণয় করি।
\(\frac{2 \sin ^{2} 63^{\circ}+1+2 \sin ^{2} 27^{\circ}}{3 \cos ^{2} 17^{\circ}-2+3 \cos ^{2} 73^{\circ}}\)
\(=\frac{2 \sin ^{2} 63^{\circ}+2 \sin ^{2}\left(90^{\circ}-63^{\circ}\right)+1}{3 \cos ^{2} 17^{\circ}+3 \cos ^{2}\left(90^{\circ}-17^{\circ}\right)-2}\)
\(=\frac{2 \sin ^{2} 63^{\circ}+2 \cos ^{2} 63^{\circ}+1}{3 \cos ^{2} 17^{\circ}+3 \sin ^{2} 17^{\circ}-2}\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta, \cos \left(90^{\circ}-\theta\right)=\sin 0\right]\)
\(=\frac{2\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)+1}{3\left(\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}\right)-2}\)
\(=\frac{2 \times 1+1}{3 \times 1-2}=\frac{2+1}{3-2}=\frac{3}{1}=3 .\)
\(\therefore\) প্রদত্ত রাশির মান \(= 3\).
\(=\frac{2 \sin ^{2} 63^{\circ}+2 \sin ^{2}\left(90^{\circ}-63^{\circ}\right)+1}{3 \cos ^{2} 17^{\circ}+3 \cos ^{2}\left(90^{\circ}-17^{\circ}\right)-2}\)
\(=\frac{2 \sin ^{2} 63^{\circ}+2 \cos ^{2} 63^{\circ}+1}{3 \cos ^{2} 17^{\circ}+3 \sin ^{2} 17^{\circ}-2}\) \(\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta, \cos \left(90^{\circ}-\theta\right)=\sin 0\right]\)
\(=\frac{2\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)+1}{3\left(\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}\right)-2}\)
\(=\frac{2 \times 1+1}{3 \times 1-2}=\frac{2+1}{3-2}=\frac{3}{1}=3 .\)
\(\therefore\) প্রদত্ত রাশির মান \(= 3\).
\({\rm{ (iv) }}\left( {\tan {1^\circ } \times \tan {2^\circ } \times \tan {3^\circ } \ldots \ldots \tan {{89}^\circ }} \right)\) এর মান নির্ণয় করি।
\(\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 89^{\circ}\)
\(=\tan 1^{\circ}+\tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ} \)
\(\quad\quad\times \ldots \times \tan 87^{\circ} \times \tan 88^{\circ} \times \tan 89^{\circ}\)
\(=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ} \times \ldots \)
\(\quad\quad\times \tan \left(90^{\circ}-3^{\circ}\right) \times \tan \left(90^{\circ}-2^{\circ}\right) \times \tan \left(90^{\circ}-1^{\circ}\right)\)
\(=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times\tan 45^{\circ} \times \cot 3^{\circ} \)
\(\quad\quad\times \cot 2^{\circ} \times \cot 1^{\circ} \quad\)[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 \(\tan \theta \times \cot \theta=1]\)
\(=\tan 45^{\circ}\)
\(= 1\)
\(=\tan 1^{\circ}+\tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ} \)
\(\quad\quad\times \ldots \times \tan 87^{\circ} \times \tan 88^{\circ} \times \tan 89^{\circ}\)
\(=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times \tan 45^{\circ} \times \ldots \)
\(\quad\quad\times \tan \left(90^{\circ}-3^{\circ}\right) \times \tan \left(90^{\circ}-2^{\circ}\right) \times \tan \left(90^{\circ}-1^{\circ}\right)\)
\(=\tan 1^{\circ} \times \tan 2^{\circ} \times \tan 3^{\circ} \times \ldots \times\tan 45^{\circ} \times \cot 3^{\circ} \)
\(\quad\quad\times \cot 2^{\circ} \times \cot 1^{\circ} \quad\)[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 \(\tan \theta \times \cot \theta=1]\)
\(=\tan 45^{\circ}\)
\(= 1\)
\({\rm{ (v) }}\sec 5A = cosec \left( {A + {{36}^\circ }} \right)\) এবং 5A ধনাত্মক সূক্ষ্মকোণ হলে, A এর মান নির্ণয় করি।
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 \(= 9^{\circ}\).
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 \(= 9^{\circ}\).
WB Board Class 10 Math Book Solution | Koshe Dekhi 24 Class 10 | পূরক কোণের ত্রিকোণমিতিক অনুপাত কষে দেখি ২৪
এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি করা বা সম্পাদনা করা নিষিদ্ধ। ভারতীয় Copywright আইন 1957 এর ধারা 63 অনুযায়ী, এই ফাইলটির সমস্ত অধিকার 'ছাত্র মিত্র Mathematics' অ্যাপ দ্বারা সংরক্ষিত। ছাত্র মিত্রের অনুমতি ছাড়া, এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি করা বা সম্পাদনা করা আইনত দন্ডনীয় অপরাধ। কেউ ছাত্র মিত্রের অনুমতি ছাড়া, এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি বা সম্পাদনা করলে ছাত্র মিত্র কতৃপক্ষ তার বিরুদ্ধে সকল প্রকার কঠোর আইনি পদক্ষেপ করবে।
West Bengal Board of Secondary Education Official Site
Class 8 : গণিত প্রভা (অষ্টম শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali.
Class 7 : গণিত প্রভা (সপ্তম শ্রেণি) বইয়ের সমস্ত সমাধান
www.wbresults.nic.in Official
Class 10 : মাধ্যমিক গণিত প্রকাশ (দশম শ্রেণি) বইয়ের সমস্ত সমাধান Class 10 Maths Solution WBBSE Bengali
Class 6 : গণিত প্রভা (ষষ্ঠ শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali
Class 9 : গণিত প্রকাশ (নবম শ্রেণি) বইয়ের সমাধান Maths Solution WBBSE Bengali
আজই Install করুন Chatra Mitra
Class 8 : গণিত প্রভা (অষ্টম শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali.
Class 7 : গণিত প্রভা (সপ্তম শ্রেণি) বইয়ের সমস্ত সমাধান
www.wbresults.nic.in Official
Class 10 : মাধ্যমিক গণিত প্রকাশ (দশম শ্রেণি) বইয়ের সমস্ত সমাধান Class 10 Maths Solution WBBSE Bengali
Class 6 : গণিত প্রভা (ষষ্ঠ শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali
Class 9 : গণিত প্রকাশ (নবম শ্রেণি) বইয়ের সমাধান Maths Solution WBBSE Bengali
আজই Install করুন Chatra Mitra
