Differentiation (Part-38) | S N De

 

$\,3(i).\,$ If $\,\,\alpha,\,\,\beta\,\,$ are positive acute angles and $\,\,(a)\,\,\sin\alpha=\frac{1}{\sqrt5},\,\,\sin\beta=\frac{1}{\sqrt{10}},\,\,$ then find $\,(\alpha+\beta).$

Sol. $\,\,\sin\alpha=\frac{1}{\sqrt5} \\ \Rightarrow \cos\alpha\\=\sqrt{1-\sin^2\alpha}\\=\sqrt{1-\frac 15}\\=\frac{2}{\sqrt5}\\ \text{Again,}\,\,\sin\beta=\frac{1}{\sqrt{10}} \\ \Rightarrow \cos\beta\\=\sqrt{1-\sin^2\beta}\\=\sqrt{1-\frac{1}{10}}\\=\frac{3}{\sqrt{10}} \\ \therefore \sin(\alpha+\beta)\\=\sin\alpha\cos\beta+\cos\alpha\sin\beta\\=\frac{1}{\sqrt5}.\frac{3}{\sqrt{10}}+\frac{2}{\sqrt5}.\frac{1}{\sqrt{10}}\\=\frac{3+2}{\sqrt{50}}\\=\frac{5}{5\sqrt2}\\=\frac{1}{\sqrt2} \\ \Rightarrow \sin(\alpha+\beta)=\frac{1}{\sqrt2}=\sin 45^{\circ} \\ \Rightarrow \alpha+\beta=45^{\circ}$

$\,\,(b)\,\,$ If $\,\tan\alpha=\cot\beta=a,\,\,$ find $\,(\alpha+\beta).$

Sol. $\,\tan\alpha=\cot\beta=a\\ \Rightarrow \tan\alpha=\cot\beta \\ \Rightarrow \tan\alpha=\tan(90^{\circ}-\beta) \\ \Rightarrow \alpha=90^{\circ}-\beta \\ \Rightarrow \alpha+\beta=90^{\circ}\quad \text{(ans.)}$

$\,(c)\,\,\,\tan\alpha=\frac 25,\,\,\tan\beta=\frac 37,\,\,$ find $\,\,(\alpha+\beta).$

Sol.  $\,\,\tan(\alpha+\beta)\\=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\=\frac{\frac 25+\frac 37}{1-\frac 25\times \frac 37}\,\,[**]\\=\frac{\frac{14+15}{35}}{1-\frac{6}{35}}\\=\frac{29/35}{29/35}\\=1 \\ \therefore  \tan(\alpha+\beta)=1=\tan 45^{\circ} \\ \Rightarrow \alpha+\beta=45^{\circ}.$ 

$\,(d)\,\,\cot\alpha=\frac 14,\,\,\cot\beta=\frac 53,\,\,$ prove that, $\,\,\alpha-\beta=45^{\circ}.$

Sol. $\,\,\cot(\alpha-\beta)\\=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}\\=\frac{\frac 14 \times \frac 53+1}{\frac 53-\frac 14}\\=\frac{\frac{5}{12}+1}{\frac{20-3}{12}}\\=\frac{17/12}{17/12}\\=1 \\ \Rightarrow \cot(\alpha-\beta)=1=\cot 45^{\circ} \\ \therefore \alpha-\beta=45^{\circ}\quad \text{(proved)}$

$\,\,3(ii)\,\,$ If $\,\,A,B,C\,$ are positive acute angles and $\,\,\tan A=\frac 47,\,\tan B=\frac 17,\, \tan C=\frac 18,\,\,$ prove that , $\,\,A+B+C=45^{\circ}.$

Sol. We have, $\,\,\tan A=\frac 47,\,\tan B=\frac 17,\, \tan C=\frac 18 \,\,\text{(Given)}$

Let $\,\,x=\tan A, y=\tan B, z=\tan C\,\,$ 

so that $\,\,\tan (A+B+C)\\=\frac{x+y+z-xyz}{1-xy-yz-zx}\\=\frac{\frac 47+\frac 17+\frac 18-\frac 47.\frac 17.\frac 18}{1-\frac 47.\frac 17-\frac 17.\frac 18-\frac 18.\frac 47}\\=\frac{4+1+\frac 78-\frac{1}{14}}{7-\frac 47-\frac 18-\frac 12}\,\,[**]\\=\frac{5+\frac{45}{56}}{7-\frac{67}{56}}\\=\frac{325/56}{325/56}\\=1 \\ \Rightarrow \tan(A+B+C)=1=\tan 45^{\circ} \\ \Rightarrow A+B+C=45^{\circ}\,\,\, \text{(proved)}$

Note [**] : Multiplying numerator and denominator by $\,\,7\,.$ 

$\,4(i).\,$ Find $\,\,\cos(\theta+\phi)\,\,$ where $\,\,\theta,\,\phi\,$are acute angles and $\,\,\sin\theta=\frac 45,\,\,\sin\phi=\frac 35.$

Sol. $\,\,\sin\theta=\frac 45 \\ \Rightarrow \cos\theta\\=\sqrt{1-\sin^2\theta}\\=\sqrt{1-\frac{16}{25}}\\=\sqrt{\frac{9}{25}}\\=\frac 35. \\ \sin\phi=\frac 35 \\ \Rightarrow \cos\phi\\=\sqrt{1-\sin^2\phi}\\=\sqrt{1-\frac{9}{25}}\\=\sqrt{\frac{16}{25}}\\=\frac 45 \\ \therefore \cos(\theta+\phi)\\=\cos\theta\cos\phi-\sin\theta\sin\phi\\=\frac 35.\frac 45-\frac 45.\frac 35\\=\frac{12}{25}-\frac{12}{25}\\=0 \quad \text{(ans.)}$

$\,\,4(ii)\,\,$ If $\,\,\sec\theta=\frac 54,\,\,\csc\phi=\frac{13}{12}\,\,$ then find the value of $\,\,\sec(\theta+\phi).$

Note : Here, by $\,\,\csc\phi\,\,$ we mean $\,\text{cosec}\,\,\phi.$ 

Sol. $\,\,\sec\theta=\frac 54 \Rightarrow \cos\theta=\frac 45\rightarrow(1) \\ \csc\phi=\frac{13}{12} \Rightarrow \sin\phi=\frac{12}{13}\rightarrow(2)$

Hence, $\,\,\sin\theta\\=\sqrt{1-\cos^2\theta}\\=\sqrt{1-\frac{16}{25}}\,\quad\text{[By (1)]}\\=\sqrt{\frac{25-16}{25}}\\=\frac 35 \\ \text{and }\,\,\cos\phi\\=\sqrt{1-\sin^2\phi}\\=\sqrt{1-\frac{144}{169}}\quad \text{[By (2)]}\\=\frac{5}{13}$

Hence, $\,\cos(\theta+\phi)\\=\cos\theta\cos\phi-\sin\theta\sin\phi\\=\frac 45.\frac{5}{13}-\frac 35.\frac{12}{13}\\=\frac{20}{65}-\frac{36}{65}\\=\frac{20-36}{65}\\=-\frac{16}{65} \\ \therefore \sec(\theta+\phi)=-\frac{65}{16}\quad \text{(ans.)}$

$\,\,4(iii)\,\,$ If $\,\,\tan\alpha=\frac 75,\,\,\tan\beta=\frac 57\,\,$ then find the value of $\,\,\tan(\alpha+\beta)\,\,$ and $\,\,\cot(\alpha-\beta).$

Sol. We have , $\,\,\tan\alpha=\frac 75\Rightarrow \cot\alpha=\frac 57 \rightarrow(1)\\ \tan\beta=\frac 57 \Rightarrow \cot\beta=\frac 75\rightarrow(2)$ 

$\,\,\tan(\alpha+\beta)\\=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\=\frac{\frac 75+\frac 57}{1-\frac 75.\frac 57}\\=\frac{\frac 75+\frac 57}{1-1}\\=\infty \\ \text{and}\,\, \cot(\alpha-\beta)\\=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}\\=\frac{\frac 57.\frac 75+1}{\frac 75-\frac 57}\quad [\text{By (1),(2)}]\\=\frac{1+1}{\frac{24}{35}}\\=2 \times \frac{35}{24}\\=\frac{35}{12}$

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