Differentiation (Part-38) | S N De

  

$\,3.\,$ Find the expansions :

$\,(i)\,\,\,\sin(A-B+C)$

Sol. $\,\,\,\sin(A-B+C)\\=\sin[(A-B)+C]\\=\sin(A-B)\cos C+\cos (A-B)\sin C\\=\left(\sin A\cos B-\cos A\sin B\right)\cos C\\+\left(\cos A\cos B+\sin A\sin B\right)\sin C\\=\cos C\sin A\cos B-\cos A\sin B\cos C\\+\sin C\cos A\cos B+\sin A\sin B\sin C$ 

$\,(ii)\,\,\,\tan(A+B-C)$

Sol. $\,\,\,\tan(A+B-C)\\=\tan[(A+B)-C]\\=\frac{\tan(A+B)-\tan C}{1+\tan(A+B)\tan C}\\=\frac{\frac{\tan A+\tan B}{1-\tan A\tan B}-\tan C}{1+\left(\frac{\tan A+\tan B}{1-\tan A\tan B}\right)\tan C}\\=\frac{\tan A+\tan B-\tan C+\tan A\tan B\tan C}{1-\tan A\tan B+\tan B\tan C+\tan C\tan A}\quad\text{(ans.)}$ 

$\,4.\,$ If $\,\,\,\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0,\,\,$ show that, $\,\,\sin(\alpha+\beta)=0;$ hence deduce that, $\,\,1+\cot\alpha\tan\beta=0.$

Sol. We have, $\,\,\,\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0 \\ \Rightarrow \cos\alpha\cos\beta-\sin\alpha\sin\beta=1\\ \Rightarrow \cos(\alpha+\beta)=1 \\ \Rightarrow \cos^2(\alpha+\beta)=1 \\ \Rightarrow 1-\sin^2(\alpha+\beta)=1 \\ \Rightarrow \sin^2(\alpha+\beta)=0 \\ \therefore \sin(\alpha+\beta)=0 \quad\text{(showed)}$

Now, $\,\,1+\cot\alpha\tan\beta\\=1+\frac{\cos\alpha}{\sin\alpha} .\frac{\sin\beta}{\cos\beta}\\=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta}\\=\frac{\sin(\alpha+\beta)}{\sin\alpha\cos\beta}\\=\frac{0}{\sin\alpha\cos\beta}\\=0.$

$\,5.\,$ If $\,\,A+B=45^{\circ},\,\,$show that, $\,(1+\tan A)(1+\tan B)=2;\,\,$ hence find the  value of $\,\,\tan(22\frac 12)^{\circ}.$

Sol. We have,  $\,\,A+B=45^{\circ}\\ \Rightarrow A+B=\frac{\pi}{4} \\ \therefore B=\frac{\pi}{4}-A\rightarrow(1)$

$\,\,(1+\tan A)(1+\tan B)\\=(1+\tan A)\left[1+\tan\left(\frac{\pi}{4}-A\right)\right]\,\,[\text{By (1)}]\\=(1+\tan A).\left[1+\frac{1-\tan A}{1+\tan A}\right]\\=(1+\tan A) \times \frac{1+\tan A+1-\tan A}{1+\tan A}\\=2\quad \text{(showed)}$

Let $\,\,A=B=\left(22\frac 12\right)^{\circ} \\ \therefore [1+\tan\left(22\frac 12\right)^{\circ}][1+\tan\left(22\frac 12\right)^{\circ}]=2 \\ \Rightarrow \left[1+\tan\left(22\frac 12\right)^{\circ}\right]^2=2 \\ \Rightarrow 1+\tan\left(22\frac 12\right)^{\circ}=\sqrt2 \\ \therefore \tan\left(22\frac 12\right)^{\circ}=\sqrt2-1.$

$\,6.\,$ If $\,\,A+B+C=\pi\,\,$ and $\,\,\cos A=\cos B\cos C,\,\,$ 

Show that, $\,\,(i)\,\,\tan A=\tan B+\tan C\\ (ii)\,\, 2\cot B\cot C=1.$

Sol. We have, $\,\,A+B+C=\pi \Rightarrow B+C=\pi-A \rightarrow(1)$ and $\,\,\cos A=\cos B\cos C\rightarrow(2)$

 $\,\,\tan B+\tan C\\=\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}\\=\frac{\sin B\cos C+\cos B\sin C}{\cos B\cos C}\\=\frac{\sin(B+C)}{\cos A}\,\,[\text{By (2)}]\\=\frac{\sin(\pi-A)}{\cos A}\,\,[\text{By (1)}]\\=\frac{\sin A}{\cos A} \\ \therefore \tan B+\tan C=\tan A.$

$\,\,(ii)\,\,2\cot B\cot C-1\\=\frac{2\cos B\cos C}{\sin B\sin C}-1\\=\frac{\cos B\cos C+\cos B\cos C-\sin B\sin C}{\sin B\sin C}\\=\frac{\cos B\cos C+\cos(B+C)}{\sin B\sin C}\\=\frac{\cos A+\cos (\pi-A)}{\sin B \sin C}\,\,[\text{By (1), (2)}]\\=\frac{\cos A-\cos A}{\sin B\sin C}\\=0 \\ \therefore 2\cot B\cot C=1.$

$\,7.\,$ An angle $\,\,\theta\,\,$ is divided in two parts $\,\,\alpha\,\,$ and $\,\,\beta\,\,$ such that $\,\,\tan\alpha :\tan \beta=x:y\,\,;\,\,$ prove that , $\,\,\sin(\alpha-\beta)=\frac{x-y}{x+y}\sin\theta.$

Sol. By question, $\,\,\theta=\alpha+\beta.$

Now,  $\,\,\frac{\tan\alpha}{\tan\beta}=\frac xy\,\,[\text{(Given)}]\\ \Rightarrow \frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta}=\frac xy \\ \Rightarrow \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta-\cos\alpha\sin\beta}=\frac{x+y}{x-y}\,\,[**] \\ \Rightarrow \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=\frac{x+y}{x-y} \\ \therefore \sin(\alpha-\beta)=\frac{x-y}{x+y}\sin\theta\,\,\text{(proved)}$

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

$\,8.\,$ If $\,\,\tan\theta=\frac{Q\sin\alpha}{P+Q\cos\alpha},\,\,$ prove that , $\,\,\tan(\alpha-\theta)=\frac{P\sin\alpha}{Q+P\cos\alpha}.$

Sol. $\,\,\tan\theta=\frac{Q\sin\alpha}{P+Q\cos\alpha}\text{(Given)}\rightarrow(1) \\ \therefore \tan(\alpha-\theta)\\=\frac{\tan\alpha-\tan\theta}{1+\tan\alpha\tan\theta}\\=\frac{\frac{\sin\alpha}{\cos\alpha}-\frac{Q\sin\alpha}{P+Q\cos\alpha}}{1+\frac{\sin\alpha}{\cos\alpha}.\frac{Q\sin\alpha}{P+Q\cos\alpha}}\,\,[\text{By (1)}]\\=\frac{P\sin\alpha+Q\sin\alpha\cos\alpha-Q\sin\alpha\cos\alpha}{P\cos\alpha+Q\cos^2\alpha+Q\sin^2\alpha}\\=\frac{P\sin\alpha}{P\cos\alpha+Q(\cos^2\alpha+\sin^2\alpha)}\\=\frac{P\sin\alpha}{P\cos\alpha+Q}\quad \text{(proved)}$

$\,9.\,$ If $\,\,\sin(\alpha+\beta)=n\sin(\alpha-\beta)\,\,$ and $\,\,n \neq -1,\,\,$ prove that, $\,\,\cot\alpha=\frac{n-1}{n+1}\cot\beta.$

Sol.  $\,\,\sin(\alpha+\beta)=n\sin(\alpha-\beta)\,\,\text{(Given)} \\ \Rightarrow \frac{\sin(\alpha+\beta)}{\sin(\alpha-\beta)}=n \\ \Rightarrow \frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{\sin(\alpha+\beta)+\sin(\alpha-\beta)}=\frac{n-1}{n+1} \\ \Rightarrow \frac{2\cos\alpha\sin\beta}{2\sin\alpha\cos\beta}=\frac{n-1}{n+1}\,\,[**]\\ \Rightarrow \frac{\cot\alpha}{\cot\beta}=\frac{n-1}{n+1} \\ \therefore \cot\alpha=\frac{n-1}{n+1}\cot\beta.$

Note [**]: $\,\,\sin(\alpha+\beta)-\sin(\alpha-\beta)\\=\sin\alpha\cos\beta+\cos\alpha\sin\beta-\sin\alpha\cos\beta\\+\cos\alpha\sin\beta\\=2\cos\alpha\sin\beta \\ \text{and}\,\,\sin(\alpha+\beta)+\sin(\alpha-\beta)\\=\sin\alpha\cos\beta+\cos\alpha\sin\beta+\sin\alpha\cos\beta\\-\cos\alpha\sin\beta\\=2\sin\alpha\cos\beta$