Differentiation (Part-38) | S N De

 

$\,\,1(i).$ Prove that, $\,\frac{\sin3\theta-\sin\theta}{\cos\theta-\cos3\theta}=\cot2\theta$

Sol. $\,\,\frac{\sin3\theta-\sin\theta}{\cos\theta-\cos3\theta}\\=\frac{2\cos\frac{3\theta+\theta}{2}\sin\frac{3\theta-\theta}{2}}{2\sin\frac{3\theta+\theta}{2}\sin\frac{3\theta-\theta}{2}}\\=\frac{\sin\theta\cos2\theta}{\sin\theta\sin2\theta}\\=\cot2\theta\,\,\text{(proved)}$

$\,\,1(ii).$ Prove that, $\,\frac{\sin A-\sin B}{\cos A+\cos B}=\tan\frac{A-B}{2}$

Sol.  $\,\frac{\sin A-\sin B}{\cos A+\cos B}\\=\frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}\\=\tan\frac{A-B}{2}\,\,\,\text{(proved)}$

$\,\,1(iii).$ Prove that, $\,\,\frac{\sin\theta+\sin2\theta+\sin3\theta}{\cos\theta+\cos2\theta+\cos3\theta}=\tan2\theta$

Sol. $\,\,\frac{\sin\theta+\sin2\theta+\sin3\theta}{\cos\theta+\cos2\theta+\cos3\theta}\\=\frac{(\sin\theta+\sin3\theta)+\sin2\theta}{(\cos\theta+\cos3\theta)+\cos2\theta}\\=\frac{2\sin\frac{3\theta+\theta}{2}\cos\frac{3\theta-\theta}{2}+\sin2\theta}{2\cos\frac{3\theta+\theta}{2}\cos\frac{3\theta-\theta}{2}+\cos2\theta}\\=\frac{2\sin2\theta\cos\theta+\sin2\theta}{2\cos2\theta\cos\theta+\cos2\theta}\\=\frac{\sin2\theta(2\cos\theta+1)}{\cos2\theta(2\cos\theta+1)}\\=\tan2\theta\,\,\,\text{(proved)}$

$1(iv)\,\,$ Prove that, $\,\,(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2\\=4\sin^2\frac{\alpha-\beta}{2}$

Sol. $\,\,(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2\\=[2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}]^2+[2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}]^2\\=4\sin^2\frac{\alpha-\beta}{2}\left(\sin^2\frac{\alpha+\beta}{2}+\cos^2\frac{\alpha+\beta}{2}\right)\\=4\sin^2\frac{\alpha-\beta}{2}\,\,\text{(proved)}$

$1(v)\,\,$ Prove that, $\,\,\cos10^{\circ}+\cos110^{\circ}+\cos130^{\circ}=0$

Sol. $\,\,\cos10^{\circ}+\cos110^{\circ}+\cos130^{\circ}\\=\cos10^{\circ}+2\cos\frac{130^{\circ}+110^{\circ}}{2}\cos\frac{130^{\circ}-110^{\circ}}{2}\\=\cos10^{\circ}+2\cos120^{\circ}\cos10^{\circ}\\=\cos10^{\circ}+2\cos(90^{\circ}+30^{\circ})\cos10^{\circ}\\=\cos10^{\circ}-2\sin30^{\circ}\cos10^{\circ}\\=\cos10^{\circ}-2.\frac 12.\cos10^{\circ}\\=\cos10^{\circ}-\cos10^{\circ}\\=0\,\,\text{(proved)}$

$1(vi)\,\,$ Prove that, $\,\sin85^{\circ}-\cos65^{\circ}=\cos55^{\circ}$

Sol. $\,\,\sin85^{\circ}-\cos65^{\circ}\\=\sin85^{\circ}-\sin(90^{\circ}-65^{\circ})\\=\sin85^{\circ}-\sin25^{\circ}\\=2\cos\frac{85^{\circ}+25^{\circ}}{2}\sin\frac{85^{\circ}-25^{\circ}}{2}\\=2\cos55^{\circ}\sin30^{\circ}\\=2\cos55^{\circ}\times \frac 12\\=\cos55^{\circ}\,\,\,\text{(proved)}$

$\,1(vii)\,$ Prove that, $\,\,\sin18^{\circ}+\cos18^{\circ}=\sqrt2\cos27^{\circ}$

Sol. $\,\,\sin18^{\circ}+\cos18^{\circ}\\=\sin18^{\circ}+\cos(90^{\circ}-72^{\circ})\\=\sin18^{\circ}+\sin72^{\circ}\\=2\sin\frac{72^{\circ}+18^{\circ}}{2}\cos\frac{72^{\circ}-18^{\circ}}{2}\\=2\sin45^{\circ}\cos27^{\circ}\\=2\times \frac{1}{\sqrt2}\times \cos27^{\circ}\\=\sqrt2\cos27^{\circ}\,\,\,\text{(proved)}$

$2.\,\,$ Prove the following :

$(i)\,\,\frac{\sin75^{\circ}+\sin15^{\circ}}{\sin75^{\circ}-\sin15^{\circ}}=\sqrt3$

Sol. $\,\,\frac{\sin75^{\circ}+\sin15^{\circ}}{\sin75^{\circ}-\sin15^{\circ}}\\=\frac{2\sin\frac{75^{\circ}+15^{\circ}}{2}\cos\frac{75^{\circ}-15^{\circ}}{2}}{2\cos\frac{75^{\circ}+15^{\circ}}{2}\sin\frac{75^{\circ}-15^{\circ}}{2}}\\=\frac{\sin45^{\circ}\cos30^{\circ}}{\cos45^{\circ}\sin30^{\circ}}\\=\frac{\frac{\sqrt3}{2}}{\frac 12}\,\,[**]\\=\sqrt3\,\,\text{(proved)}$

Note[**] : $\,\,\sin45^{\circ}=\cos45^{\circ}=\frac{1}{\sqrt2}$

$(ii)\,\,\frac{\cos45^{\circ}-\cos75^{\circ}}{\sin45^{\circ}+\sin75^{\circ}}=2-\sqrt3$

Sol. $\,\,\frac{\cos45^{\circ}-\cos75^{\circ}}{\sin45^{\circ}+\sin75^{\circ}}\\=\frac{2\sin\frac{75^{\circ}+45^{\circ}}{2}\sin\frac{75^{\circ}-45^{\circ}}{2}}{2\sin\frac{75^{\circ}+45^{\circ}}{2}\cos\frac{75^{\circ}-45^{\circ}}{2}}\\=\frac{\sin60^{\circ}\sin15^{\circ}}{\sin60^{\circ}\cos15^{\circ}}\\=\tan15^{\circ}\\=2-\sqrt3$

Note : $\,\,\tan 15^{\circ}\\=\tan(45^{\circ}-30^{\circ})\\=\frac{\tan45^{\circ}-\tan30^{\circ}}{1+\tan45^{\circ}\tan30^{\circ}}\\=\frac{1-1/\sqrt3}{1+1/\sqrt3},\,\,[\tan45^{\circ}=1]\\=\frac{\sqrt3-1}{\sqrt3+1}\\=\frac{(\sqrt3-1)(\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)}\\=\frac{(\sqrt3-1)^2}{(\sqrt3)^2-1^2}\\=\frac 12 \times (3+1-2\sqrt3)\\=\frac 12 \times (4-2\sqrt3)\\=2-\sqrt3$

$(iii)\,\,\frac{\cos20^{\circ}-\sin20^{\circ}}{\cos20^{\circ}+\sin20^{\circ}}=\tan25^{\circ}$

Sol. $\,\,\frac{\cos20^{\circ}-\sin20^{\circ}}{\cos20^{\circ}+\sin20^{\circ}}\\=\frac{\cos20^{\circ}-\cos(90^{\circ}-20^{\circ})}{\cos20^{\circ}+\cos(90^{\circ}-20^{\circ})}\\=\frac{\cos20^{\circ}-\cos70^{\circ}}{\cos20^{\circ}+\cos70^{\circ}}\\=\frac{2\sin\frac{20^{\circ}+70^{\circ}}{2}\sin\frac{70^{\circ}-20^{\circ}}{2}}{2\cos\frac{70^{\circ}+20^{\circ}}{2}\cos\frac{70^{\circ}-20^{\circ}}{2}}\\=\frac{\sin45^{\circ}\sin25^{\circ}}{\cos45^{\circ}\cos25^{\circ}}\\=\frac{\sin25^{\circ}}{\cos25^{\circ}}\,\,[**]\\=\tan25^{\circ}\,\,\text{(proved)}$

Note[**]: $\,\,\sin45^{\circ}=\cos45^{\circ}=\frac{1}{\sqrt2}$

$(iv)\,\,\frac{\cos10^{\circ}-\sin10^{\circ}}{\cos10^{\circ}+\sin10^{\circ}}=\tan 35^{\circ}$

Sol. $\,\,\frac{\cos10^{\circ}-\sin10^{\circ}}{\cos10^{\circ}+\sin10^{\circ}}\\=\frac{\cos10^{\circ}-\cos80^{\circ}}{\cos10^{\circ}+\cos80^{\circ}}\,\,[*]\\=\frac{2\sin\frac{80^{\circ}+10^{\circ}}{2}\sin\frac{80^{\circ}-10^{\circ}}{2}}{2\cos\frac{80^{\circ}+10^{\circ}}{2}\cos\frac{80^{\circ}-10^{\circ}}{2}}\\=\frac{\sin45^{\circ}\sin35^{\circ}}{\cos45^{\circ}\cos35^{\circ}}\\=\frac{\frac{1}{\sqrt2}\sin35^{\circ}}{\frac{1}{\sqrt2}\cos35^{\circ}}\\=\tan35^{\circ}\,\,\text{(proved)}$

Note[*] : $\sin10^{\circ}=\cos(90^{\circ}-10^{\circ})=\cos80^{\circ}$

$(v)\,\,\frac{\cos2\alpha-\cos2\beta}{\sin2\alpha+\sin2\beta}=\tan(\beta-\alpha)$

Sol. $\,\frac{\cos2\alpha-\cos2\beta}{\sin2\alpha+\sin2\beta}\\=\frac{2\sin\frac{2(\alpha+\beta)}{2}\sin\frac{2(\beta-\alpha)}{2}}{2\sin\frac{2(\alpha+\beta)}{2}\cos\frac{2(\alpha-\beta)}{2}}\\=\frac{\sin(\beta-\alpha)}{\cos(\beta-\alpha)}\\=\tan(\beta-\alpha)\,\,\text{(proved)}$

$3.\,\,$ If $\,\,x\cos\alpha+y\sin\alpha=x\cos\beta+y\sin\beta,\,\,$ show that, $\,y=x\tan\frac{\alpha+\beta}{2}$

sol. $\,\,x\cos\alpha+y\sin\alpha=x\cos\beta+y\sin\beta \\ \Rightarrow y(\sin\alpha-\sin\beta)=x(\cos\beta-\cos\alpha)\\ \Rightarrow y=\frac{x(\cos\beta-\cos\alpha)}{(\sin\alpha-\sin\beta)} \\ \Rightarrow y=\frac{x \times 2\sin\frac{\beta+\alpha}{2}\sin\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}\\ \therefore y=x\tan\frac{\alpha+\beta}{2}$

$4.\,\,$ If $\,\cos\alpha+\cos\beta=\frac13,\,\,\sin\alpha+\sin\beta=\frac14,\,$show that, $\,\,\,\,\tan\frac{\alpha+\beta}{2}=\frac34.$

Sol. We have, $\,\cos\alpha+\cos\beta=\frac13,\rightarrow(1)\\ \sin\alpha+\sin\beta=\frac14\rightarrow(2)$

From (1) and (2), we get $\,\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}=\frac34 \\ \Rightarrow \frac{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}=\frac 34 \\ \Rightarrow \tan\frac{\alpha+\beta}{2}=\frac 34\,\,\text{(proved)}$

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