Differentiation (Part-38) | S N De

 

$\,1(i).\,$ Prove that, $\,\,\,\frac{\sin\alpha}{1-\cos\alpha}=\cot\frac{\alpha}{2}$

Sol. $\,\,\,\,\,\frac{\sin\alpha}{1-\cos\alpha}\\=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\sin^2\frac{\alpha}{2}}\\=\frac{\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}}\\=\cot\frac{\alpha}{2}$

$\,1(ii).\,$ Prove that, $\,\,\,\frac{\sin2\theta}{1+\cos2\theta}.\frac{\cos\theta}{1+\cos\theta}=\tan\frac{\theta}{2}$

Sol. $\,\,\,\frac{\sin2\theta}{1+\cos2\theta}.\frac{\cos\theta}{1+\cos\theta}\\=\frac{2\sin\theta\cos\theta}{2\cos^2\theta}.\frac{\cos\theta}{2\cos^2\frac{\theta}{2}}\\=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\\=\frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}\\=\tan\frac{\theta}{2}$

$\,1(iii).\,$ Prove that, $\,\,\,\tan\frac{A}{2}-\cot\frac A2+2\cot A=0$

Sol. $\,\,\,\tan\frac{A}{2}-\cot\frac A2+2\cot A\\=\frac{\sin\frac A2}{\cos\frac A2}-\frac{\cos\frac A2}{\sin\frac A2}+2\cot A\\=\frac{\sin^2\frac A2-\cos^2\frac A2}{\sin\frac A2\cos\frac A2}+2\cot A\\=\frac{-2(\cos^2\frac A2-\sin^2\frac A2)}{2\sin\frac A2\cos\frac A2}+2\cot A\\=\frac{-2\cos A}{\sin A}+2\cot A\\=-2\cot A+2\cot A\\=0$

$\,1(iv).\,$ Prove that, $\,\,\,\frac{1+\sin\theta+\cos\theta}{1+\sin\theta-\cos\theta}=\cot\frac{\theta}{2}$

Sol. $\,\,\,\frac{1+\sin\theta+\cos\theta}{1+\sin\theta-\cos\theta}\\=\frac{\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2})}\\=\frac{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})^2+(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})^2-(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}\\=\frac{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}+\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}-\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}\\=\frac{2\cos\frac{\theta}{2}}{2\sin\frac{\theta}{2}}\\=\cot\frac{\theta}{2}$

Note : $\,\,1+\sin\theta=(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2})+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ \text{and}\,\,\cos\theta=\cos(2.\frac{\theta}{2})=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$

$\,1(v).\,$ Prove that, $\,\,\,2\csc \theta=\tan\frac{\theta}{2}+\cot\frac{\theta}{2}$

Note : By $\,\csc \theta,\,\,$ we mean $\,\text{cosec}\,\,\theta.$

Sol. $\,\,\tan\frac{\theta}{2}+\cot\frac{\theta}{2}\\=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}+\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\\=\frac{\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{1}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{2}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{2}{\sin\theta}\\=2\csc\theta$ 

$\,1(vi).\,$ Prove that, $\,\,\, \sec\alpha-\tan\alpha=\cot\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)$

Sol. We know, $\,\cot(A+B)=\frac{\cot A \cot B-1}{\cot B+\cot A} $

and so,$\,\,\cot\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\\=\frac{\cot\frac{\alpha}{2}-1}{1+\cot\frac{\alpha}{2}},\,\,[\text{Since,}\,\,\cot\frac{\pi}{4}=1]\\=\frac{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}\,\,[**]\\=\frac{(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})}{(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})}\\=\frac{(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2})^2}{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}\\=\frac{\cos^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}-2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos \alpha}\\=\frac{1-\sin\alpha}{\cos\alpha}\\=\frac{1}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}\\=\sec\alpha-\tan\alpha$

Note[**] : Multiplying numerator and denominator by $\,\sin\frac{\alpha}{2}.$

$\,1(vii).\,$ Prove that, $\,\,\, \frac{2\sin\theta-\sin2\theta}{2\sin\theta+\sin2\theta}=\tan^2\frac{\theta}{2}$

Sol.  $\,\,\, \frac{2\sin\theta-\sin2\theta}{2\sin\theta+\sin2\theta}\\=\frac{2\sin\theta-2\sin\theta\cos\theta}{2\sin\theta+2\sin\theta\cos\theta}\\=\frac{2\sin\theta(1-\cos\theta)}{2\sin\theta(1+\cos\theta)}\\=\frac{1-\cos\theta}{1+\cos\theta}\\=\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\\=\tan^2\frac{\theta}{2}$

$\,1(viii).\,$ Prove that, $\,\,\, (\sin\theta+\sin\phi)^2+(\cos\theta+\cos\phi)^2=4\cos^2\frac{\theta-\phi}{2}$

Sol. $\,\,\, (\sin\theta+\sin\phi)^2+(\cos\theta+\cos\phi)^2\\=\sin^2\theta+\sin^2\phi+2\sin\theta\sin\phi\\+\cos^2\theta+\cos^2\phi+2\cos\theta\cos\phi\\=(\sin^2\theta+\cos^2\theta)+(\sin^2\phi+\cos^2\phi)\\+2(\cos\theta\cos\phi+\sin\theta\sin\phi)\\=1+1+2\cos(\theta-\phi)\\=2+2\cos(\theta-\phi)\\=2\left(1+\cos(\theta-\phi)\right)\\=2.2\cos^2\frac{\theta-\phi}{2}\\=4\cos^2\frac{\theta-\phi}{2}$

$\,1(ix).\,$ Prove that, $\,\,\,(\sin\theta-\sin\phi)^2+(\cos\theta-\cos\phi)^2=4\sin^2\frac{\theta-\phi}{2} $

Sol. $\,\,\,(\sin\theta-\sin\phi)^2+(\cos\theta-\cos\phi)^2\\=\sin^2\theta+\sin^2\phi-2\sin\theta\sin\phi\\+\cos^2\theta+\cos^2\phi-2\cos\theta\cos\phi\\=(\sin^2\theta+\cos^2\theta)+(\sin^2\phi+\cos^2\phi)\\-2(\cos\theta\cos\phi+\sin\theta\sin\phi)\\=1+1-2\cos(\theta-\phi)\\=2\left[1-\cos(\theta-\phi)\right]\\=2.2\sin^2\frac{\theta-\phi}{2}\\=4\sin^2\frac{\theta-\phi}{2} $

$\,1(x).\,$ Prove that, $\,\,\, \frac{\cos\frac{\alpha}{2}-\sqrt{1+\sin\alpha}}{\sin\frac{\alpha}{2}-\sqrt{1+\sin\alpha}}=\tan\frac{\alpha}{2}$

Sol.  $\,\,\, \frac{\cos\frac{\alpha}{2}-\sqrt{1+\sin\alpha}}{\sin\frac{\alpha}{2}-\sqrt{1+\sin\alpha}}\\=\frac{\cos\frac{\alpha}{2}-\sqrt{\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}+2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}}{\sin\frac{\alpha}{2}-\sqrt{\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}+2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}}\\=\frac{\cos\frac{\alpha}{2}-(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2})}{\sin\frac{\alpha}{2}-(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2})}\\=\frac{-\sin\frac{\alpha}{2}}{-\cos\frac{\alpha}{2}}\\=\tan\frac{\alpha}{2}$

$\,1(xi).\,$ Prove that, $\,\,\, 2\sin\frac{\pi}{8}=\sqrt{2-\sqrt2}$

Sol. $\,\cos\frac{\pi}{4}=1-2\sin^2\frac{\pi}{8}\,\,[**]\\ \Rightarrow \frac{1}{\sqrt2}=1-2\sin^2\frac{\pi}{8} \\ \Rightarrow 2\sin^2\frac{\pi}{8}=1-\frac{1}{\sqrt2}\\~~~~~~~~~~~~~~~~~~=\frac{\sqrt2-1}{\sqrt2}\\~~~~~~~~~~~~~~~~~~=\frac{2-\sqrt2}{2}\\ \Rightarrow 4\sin^2\frac{\pi}{8}=2-\sqrt2 \\ \Rightarrow 2\sin\frac{\pi}{8}=\sqrt{2-\sqrt2}$

Note [**] : We have used the formula : $\,\,\cos\theta=1-2\sin^2\frac{\theta}{2}$

$\,1(xii).\,$ Prove that, $\,\,\, 2\cos\frac{\pi}{8}=\sqrt{2+\sqrt2}$

Sol. $\,\cos\frac{\pi}{4}=2\cos^2\frac{\pi}{8}-1\,\,[**]\\ \Rightarrow \frac{1}{\sqrt2}=2\cos^2\frac{\pi}{8}-1 \\ \Rightarrow 2\cos^2\frac{\pi}{8}=1+\frac{1}{\sqrt2}\\~~~~~~~~~~~~~~~~~~=\frac{\sqrt2+1}{\sqrt2}\\~~~~~~~~~~~~~~~~~~=\frac{2+\sqrt2}{2}\\ \Rightarrow 4\cos^2\frac{\pi}{8}=2+\sqrt2 \\ \Rightarrow 2\cos\frac{\pi}{8}=\sqrt{2+\sqrt2}$

Note [**] : We have used the formula : $\,\,\cos\theta=2\cos^2\frac{\theta}{2}-1$

$\,1(xiii).\,$ Prove that, $\,\,\, \cot^236^{\circ}\cot^272^{\circ}=\frac 15$

Sol. $\,\,\, \cot^236^{\circ}\cot^272^{\circ}\\=\frac{\cos^236^{\circ}}{\sin^236^{\circ}} \times \frac{\cos^272^{\circ}}{\sin^272^{\circ}}\\=\frac{\left(\frac{\sqrt5+1}{4}\right)^2}{\left(\frac{\sqrt{10-2\sqrt5}}{4}\right)^2} \times \frac{\left(\frac{\sqrt5-1}{4}\right)^2}{\left(\frac{\sqrt{10+2\sqrt5}}{4}\right)^2}\\=\frac{(\sqrt5+1)^2}{\left(\sqrt{10-2\sqrt5}\right)^2} \times \frac{(\sqrt5-1)^2}{\left(\sqrt{10+2\sqrt5}\right)^2}\\=\frac{5+2\sqrt5+1}{10-2\sqrt5}\times \frac{5-2\sqrt5+1}{10+2\sqrt5}\\=\frac{2(3+\sqrt5)}{2(5-\sqrt5)} \times\frac{2(3-\sqrt5)}{2(5+\sqrt5)}\\=\frac{(3+\sqrt5)(3-\sqrt5)}{(5-\sqrt5)(5+\sqrt5)}\\=\frac{3^2-(\sqrt5)^2}{5^2-(\sqrt5)^2}\\=\frac{9-5}{25-5}\\=\frac{4}{20}\\=\frac 15$

$\,1(xiv).\,$ Prove that, $\,\,\, \cos\frac{\pi}{5}-\cos\frac{2\pi}{5}=\frac 12$

Sol. $\,\,\, \cos\frac{\pi}{5}-\cos\frac{2\pi}{5}\\=\cos36^{\circ}-\cos72^{\circ}\\=-2\sin\frac{72^{\circ}+36^{\circ}}{2}\sin\frac{36^{\circ}-72^{\circ}}{2}\\=-2\sin 54^{\circ}\sin(-18^{\circ})\\=2\sin54^{\circ}\sin18^{\circ}\\=2 \times \frac{\sqrt5+1}{4} \times \frac{\sqrt5-1}{4}\\=\frac 18 \times (5-1)\,\,[**]\\=\frac 12$

Note [**] : $\,(\sqrt5+1)(\sqrt5-1)=(\sqrt5)^2-1^2=5-1=4$

$\,1(xv).\,$ Prove that, $\,\,\, \cot70^{\circ}+4\cos70^{\circ}=\sqrt3$

Sol. $\,\,\,\,\,\,\,\,\cot 70^{\circ} + 4 \cos 70^{\circ} \\= \frac{\cos 70^{\circ}}{\sin 70^{\circ}} + 4 \cos 70^{\circ}\\=\frac{\cos 70^{\circ}+4\sin 70^{\circ}\cos70^{\circ}}{\sin 70^{\circ}}\\=\frac{\cos 70^{\circ}+2(2\sin 70^{\circ}\cos70^{\circ})}{\sin 70^{\circ}} \\=\frac{\cos(90^{\circ}-20^{\circ})+2\sin(2 \times 70^{\circ})}{\sin 70^{\circ}}\\=\frac{\sin 20^{\circ}+2\sin 140^{\circ}}{\sin(90^{\circ}-20^{\circ})}\\=\frac{\sin 20^{\circ}+\sin(180^{\circ}-40^{\circ})}{\cos 20^{\circ}}\\=\frac{\sin 20^{\circ}+2\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{\sin 20^{\circ}+\sin 40^{\circ}+\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{2\sin\frac{40^{\circ}+20^{\circ}}{2}\cos\frac{40^{\circ}-20^{\circ}}{2}+\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{2\sin 30^{\circ}\cos 10^{\circ}+\sin 40^{\circ}}{\cos20^{\circ}}\\=\frac{2\times \frac 12 \times \cos10^{\circ}+\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{\cos10^{\circ}+\sin 40^{\circ}}{\cos20^{\circ}}\\=\frac{\cos(90^{\circ}-80^{\circ})+\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{\sin 80^{\circ}+\sin 40^{\circ}}{\cos 20^{\circ}}\\=\frac{2\sin\frac{80^{\circ}+40^{\circ}}{2}\cos\frac{80^{\circ}-40^{\circ}}{2}}{\cos 20^{\circ}}\\=\frac{2\sin 60^{\circ}\cos 20^{\circ}}{\cos 20^{\circ}}\\=2\sin 60^{\circ}\\=2\times \frac{\sqrt3}{2}\\=\sqrt3$