Differentiation (Part-38) | S N De

 

$\,5.\,$ If $\,\,\tan\frac{\theta}{2}=\tan^3\frac{\phi}{2}\,\,\text{and}\,\,\tan\phi=2\tan\alpha,\,\,$ prove that, $\,\,\phi+\theta=2\alpha.$

Sol. $\,\,\tan\left(\frac{\theta+\phi}{2}\right)\\=\frac{\tan\frac{\phi}{2}+\tan\frac{\theta}{2}}{1-\tan\frac{\phi}{2}\tan\frac{\theta}{2}}\\=\frac{\tan\frac{\phi}{2}+\tan^3\frac{\phi}{2}}{1-\tan\frac{\phi}{2}\tan^3\frac{\phi}{2}}\,\,[\,\text{Since,}\,\tan\frac{\theta}{2}=\tan^3\frac{\phi}{2}]\\=\frac{\tan\frac{\phi}{2}\left(1+\tan^2\frac{\phi}{2}\right)}{\left(1+\tan^2\frac{\phi}{2}\right)\left(1-\tan^2\frac{\phi}{2}\right)}\\=\frac 12\times \frac{2\tan\frac{\phi}{2}}{1-\tan^2\frac{\phi}{2}}\\=\frac 12 \tan(2\times \frac{\phi}{2})\\=\frac 12\tan\phi\\=\tan\alpha\,\,[\,\tan\phi=2\tan\alpha\,\text{(given)}] \\ \Rightarrow \frac{\theta+\phi}{2}=\alpha \\ \Rightarrow \theta+\phi=2\alpha.\,\,\text{(proved)}$

$\,6.\,$ If $\,\cos\theta=\frac{a\cos\phi+b}{a+b\cos\phi},\,$ show that, $\,\,\tan\frac{\theta}{2}=\pm\sqrt{\frac{a-b}{a+b}}\tan\frac{\phi}{2}$

Sol.  $\,\,\tan^2\frac{\theta}{2}\\=\frac{1-\cos\theta}{1+\cos\theta}\\=\frac{1-\frac{a\cos\phi+b}{a+b\cos\phi}}{1+\frac{a\cos\phi+b}{a+b\cos\phi}}\\=\frac{a+b\cos\phi-a\cos\phi-b}{a+b\cos\phi+a\cos\phi+b}\\=\frac{(a-b)-(a-b)\cos\phi}{(a+b)+(a+b)\cos\phi}\\=\frac{a-b}{a+b}.\frac{1-\cos\phi}{1+\cos\phi}\\=\frac{a-b}{a+b}.\tan^2\frac{\phi}{2} \\ \therefore\,\,\tan\frac{\theta}{2}=\pm\sqrt{\frac{a-b}{a+b}}\tan\frac{\phi}{2}\,\,\text{(proved)}$

$\,7.\,$ If $\,\,\sin\theta=\frac{a-b}{a+b},\,\,$ show that, $\,\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\pm\sqrt{\frac ba}$

Sol. $\,\,\tan^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\\=\left[\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}\right]^2\\=\frac{\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}{\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2}\\=\frac{1-\sin\theta}{1+\sin\theta}\,\,[**]\\=\frac{1-\frac{a-b}{a+b}}{1+\frac{a-b}{a+b}}\,\,\,[\sin\theta=\frac{a-b}{a+b}\,\text{(given)}]\\=\frac{2b}{2a}\\=\frac ba \\ \Rightarrow \tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\pm\sqrt{\frac ba}$

Note[**] : $\,\,\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2\\=\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\=1-\sin(2.\frac{\theta}{2})\\=1-\sin\theta \\ \text{Similarly,}\,\,\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)^2\\=\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\=1+\sin(2.\frac{\theta}{2})\\=1+\sin\theta$

$\,8.\,$ If $\,\,x\cos\alpha+y\sin\alpha=x\cos\beta+y\sin\beta=k,\,\,$ show that, $\,\,\frac{x}{\cos\frac{\alpha+\beta}{2}}=\frac{y}{\sin\frac{\alpha+\beta}{2}}=\frac{k}{\cos\frac{\alpha-\beta}{2}}$

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Sol. $\,\,x\cos\alpha+y\sin\alpha=k \\ \Rightarrow x\cos\alpha+y\sin\alpha-k=0 \cdots(1) \\ \text{and}\,\,x\cos\beta+y\sin\beta=k \\ \Rightarrow x\cos\beta+y\sin\beta-k=0 \cdots(2) $

From (1) and (2), we get 

$\,\frac{x}{-k\sin\alpha+k\sin\beta}=\frac{y}{-k\cos\beta+k\cos\alpha}=\frac{1}{\cos\alpha\sin\beta-\sin\alpha\cos\beta} \\ \Rightarrow \frac{x}{-k.2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}}=\frac{y}{-k.2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}=\frac{1}{-\sin(\alpha-\beta)} \\ \Rightarrow \frac{x}{2k\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}}=\frac{y}{2k\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}}=\frac{1}{2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha-\beta}{2}} \\ \therefore \frac{x}{\cos\frac{\alpha+\beta}{2}}=\frac{y}{\sin\frac{\alpha+\beta}{2}}=\frac{k}{\cos\frac{\alpha-\beta}{2}}$

$\,9.\,$ If $\,\,\tan\frac{\alpha}{2}=\frac{1}{\sqrt3}\tan\frac{\beta}{2},\,$ show that $\,\,\cos\beta=\frac{2\cos\alpha-1}{2-\cos\alpha}.$

Sol. We have, $\,\,\tan\frac{\alpha}{2}=\frac{1}{\sqrt3}\tan\frac{\beta}{2}\cdots(1)$

Now, $\,\quad\cos\beta\\=\frac{1-\tan^2\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}} \\=\frac{1-3\tan^2\frac{\alpha}{2}}{1+3\tan^2\frac{\alpha}{2}}\,[\text{By (1)}]\\=\frac{\cos^2\frac{\alpha}{2}-3\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}+3\sin^2\frac{\alpha}{2}}\\=\frac{2\left(\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}\right)-\left(\cos^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}\right)}{\left(\cos^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}\right)+2\sin^2\frac{\alpha}{2}}\\=\frac{2\cos\alpha-1}{1+(1-\cos\alpha)}\\=\frac{2\cos\alpha-1}{2-\cos\alpha}\,\,\text{(proved)}$