Differentiation (Part-38) | S N De

 

$\,10(i).\,$ If $\,\,\cot\alpha\cot\beta=3,\,\,$ show that , $\,\,\frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=2.$

Sol. $\,\,\cot\alpha\cot\beta=3 \\ \Rightarrow \frac{\cos\alpha\cos\beta}{\sin\alpha\sin\beta}=\frac 31 \\ \Rightarrow \frac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\frac{3+1}{3-1} \,\,[**]\\ \Rightarrow \frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=\frac{4}{2}=2\quad\text{(showed)}$

Note[**] : By componendo and dividendo formula. 

$\,10(ii).\,$ Find the value of $\,\,\frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)},\,\,$ given that $\,\,\tan\alpha=2\tan\beta.$

Sol. $\quad \tan\alpha=2\tan\beta\,\,\,\text{(Given)} \\ \Rightarrow \frac{\sin\alpha}{\cos\alpha}=2\frac{\sin\beta}{\cos\beta} \\ \Rightarrow \frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta}=2 \\ \Rightarrow \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta-\cos\alpha\sin\beta}=\frac{2+1}{2-1}\,\,[**] \\ \Rightarrow \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=\frac 31=3\quad\text{(ans.)}$

Note[**] : By componendo and dividendo formula. 

$\,10(iii)\,\,$ If $\,\,0^{\circ}<\theta<90^{\circ}\,\,\,$ and $\,\,\,\cos\theta+\sin\theta=\sqrt2,\,\,$ find the value of $\,\,\cos3\theta.$

Sol. $\,\,\cos\theta+\sin\theta=\sqrt2 \\ \Rightarrow \frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta=1 \\ \Rightarrow \sin45^{\circ}\cos\theta+\cos45^{\circ}\sin\theta=1 \\ \Rightarrow \sin(45^{\circ}+\theta)=1=\sin90^{\circ} \\ \Rightarrow 45^{\circ}+\theta=90^{\circ} \\ \therefore \theta=90^{\circ}-45^{\circ}=45^{\circ}.$

Now, $\,\,\cos3\theta\\=\cos(3\times 45^{\circ})\\=\cos(90^{\circ}+45^{\circ})\\=-\sin45^{\circ}\\=-\frac{1}{\sqrt2}.$

$\,11.\,$ If $\,\,\tan\alpha=\frac{x\sin\beta}{1-x\cos\beta}\,\,$ and $\,\,\tan\beta=\frac{y\sin\alpha}{1-y\cos\alpha},\,\,$ show that , $\,\,\frac{\sin\alpha}{\sin\beta}=\frac xy.$

Sol.  $\,\,\tan\alpha=\frac{x\sin\beta}{1-x\cos\beta}\\ \Rightarrow \frac{\sin\alpha}{\cos\alpha}=\frac{x\sin\beta}{1-x\cos\beta} \\ \Rightarrow \sin\alpha(1-x\cos\beta)=\cos\alpha(x\sin\beta) \\ \Rightarrow \sin\alpha-x\sin\alpha\cos\beta=x\cos\alpha\sin\beta \\ \Rightarrow \sin\alpha=x(\sin\alpha\cos\beta+\cos\alpha\sin\beta) \\ \Rightarrow \sin\alpha=x\sin(\alpha+\beta)\rightarrow(1)$

Again, $\,\,\tan\beta=\frac{y\sin\alpha}{1-y\cos\alpha}\\ \Rightarrow \frac{\sin\beta}{\cos\beta}=\frac{y\sin\alpha}{1-y\cos\alpha} \\ \Rightarrow \sin\beta(1-y\cos\alpha)=y\sin\alpha\cos\beta \\ \Rightarrow \sin\beta=y(\sin\alpha\cos\beta+\cos\alpha\sin\beta) \\ \Rightarrow \sin\beta=y\sin(\alpha+\beta)\rightarrow(2)$

Hence, from $\,\,(1),(2)\,\,$ we get, $\,\,\frac{\sin\alpha}{\sin\beta}=\frac xy\quad \text{(showed)}$

$\,12.\,$ If $\,\,\tan\theta+\tan\phi=x\,\,$ and $\,\,\cot\theta+\cot\phi=y\,,\,\,$ prove that, $\,\,\cot(\theta+\phi)=\frac 1x-\frac 1y.$

Sol. $\,\,\tan\theta+\tan\phi=x \\ \Rightarrow \frac{\sin\theta}{\cos\theta}+\frac{\sin\phi}{\cos\phi}=x \\ \Rightarrow \frac{\sin\theta\cos\phi+\cos\theta\sin\phi}{\cos\phi\cos\theta}=x \\ \Rightarrow \frac{\sin(\theta+\phi)}{\cos\theta\cos\phi}=x \\ \Rightarrow  \frac{\cos\theta\cos\phi}{\sin(\theta+\phi)}=\frac 1x \rightarrow(1)$

Again, $\,\,\cot\theta+\cot\phi=y\\ \Rightarrow \frac{\cos\theta}{\sin\theta}+\frac{\cos\phi}{\sin\phi}=y \\ \Rightarrow \frac{\sin\phi\cos\theta+\cos\phi\sin\theta}{\sin\theta\sin\phi}=y \\ \Rightarrow \frac{\sin(\theta+\phi)}{\sin\theta\sin\phi}=y \\ \Rightarrow \frac{\sin\theta\sin\phi}{\sin(\theta+\phi)}=\frac 1y\rightarrow(2)$

From $\,\,(1),(2)\,\,$ we get, $\,\,\frac 1x-\frac 1y=\frac{\cos\theta\cos\phi-\sin\theta\sin\phi}{\sin(\theta+\phi)} \\ \Rightarrow \frac 1x-\frac 1y=\frac{\cos(\theta+\phi)}{\sin(\theta+\phi)} \\ \therefore \cot(\theta+\phi)=\frac 1x-\frac 1y\quad \text{(proved)}$

$\,13.\,$ If $\,\,\tan\theta=\frac{2x-k}{k\sqrt3}\,\,$ and $\,\,\tan\psi=\frac{x\sqrt3}{2k-x}\,,$ then show that , $|\psi-\theta|=30^{\circ}.$

Sol. We have , $\,\,\tan\theta=\frac{2x-k}{k\sqrt3}\rightarrow(1)$ and $\,\,\tan\psi=\frac{x\sqrt3}{2k-x}\rightarrow (2)$

Now, $\,\,\tan(\theta+30^{\circ})\\=\frac{\tan\theta+\tan 30^{\circ}}{1-\tan\theta\tan 30^{\circ}}\\=\frac{\tan\theta+\frac{1}{\sqrt3}}{1-\tan\theta.\frac{1}{\sqrt3}}\\=\frac{\frac{2x-k}{k\sqrt3}+\frac{1}{\sqrt3}}{1-\frac{2x-k}{k\sqrt3}.\frac{1}{\sqrt3}}\,\,\text{[By (1)]}\\=\frac{2x-k+k}{k\sqrt3} \times \frac{3k}{3k-2x+k}\\=\frac{2x\sqrt3}{2(2k-x)}\\=\frac{x\sqrt3}{2k-x}\\=\tan\psi\,\,[\text{By (2)}] \\ \Rightarrow \theta+30^{\circ}=\psi \\ \therefore \theta-\psi=-30^{\circ} \\ \therefore |\theta-\psi|=|-30^{\circ}|=30^{\circ}\quad\text{(showed)}$