Differentiation (Part-38) | S N De

 

$\,5.\,$ If $\,\,A\,$ and $\,B\,$ are positive acute angles and $\,\,\sin A=\frac 35,\,\,\cos B=\frac{12}{13},\,\,$ find the value of $\,\,\frac{\tan A-\tan B}{1+\tan A\tan B}.$

Sol. We have, $\,\sin A=\frac 35 \\ \therefore \cos A\\=\sqrt{1-\sin^2A}\\=\sqrt{1-(\frac 35)^2}\\=\sqrt{1-\frac{9}{25}}\\=\sqrt{\frac{25-9}{25}}\\=\sqrt{\frac{16}{25}}\\=\frac{4}{5}\\ \therefore  \tan A=\frac{\sin A}{\cos A}=\frac{3/5}{4/5}=\frac 34 \rightarrow(1)$

Similarly, $\,\cos B=\frac{12}{13}\,\,\text{(given)} \\ \therefore \sin B\\=\sqrt{1-\cos^2B}\\=\sqrt{1-(\frac{12}{13})^2}\\=\sqrt{1-\frac{144}{169}}\\=\sqrt{\frac{169-144}{169}}\\=\sqrt{\frac{25}{169}}\\=\frac{5}{13}\\ \therefore  \tan B=\frac{\sin B}{\cos B}=\frac{5/13}{12/13}=\frac{5}{12} \rightarrow(2)$

So, $\,\,\frac{\tan A-\tan B}{1+\tan A\tan B}\\=\frac{\frac 34-\frac{5}{12}}{1+\frac 34 \times \frac{5}{12}}\,\,\,[\text{By (1),(2)}]\\=\frac{\frac{9-5}{12}}{1+\frac{5}{16}}\\=\frac{1/3}{21/16}\\=\frac 13 \times \frac{16}{21}\\=\frac{16}{63}.$

$\,6.\,$ If $\,\,\tan\theta=\frac ab,\,\,$ find the value of $\,\,\frac{b\cos\theta-a\sin\theta}{b\cos\theta+a\sin\theta}.$

Sol. We have , $\,\,\,\tan\theta=\frac ab \\ \Rightarrow \frac{\sin\theta}{\cos\theta}=\frac ab \\ \Rightarrow \frac{\sin\theta}{a}=\frac{\cos\theta}{b}=k(\neq 0),\text{(constant)} \\ \therefore \sin\theta=ak,\,\,\cos\theta=bk\rightarrow(1) $

So, $\,\,\frac{b\cos\theta-a\sin\theta}{b\cos\theta+a\sin\theta}\\=\frac{b.bk-a.ak}{b.bk+a.ak}\quad[\text{By (1)}]\\=\frac{k(b^2-a^2)}{k(b^2+a^2)}\\=\frac{b^2-a^2}{b^2+a^2}.$

$\,7.\,$ If $\,\,\cos\theta+\sec\theta=\sqrt3,\,\,$ show that, $\,\cos^3\theta+\sec^3\theta=0.$

Sol. $\,\,\cos\theta+\sec\theta=\sqrt3\rightarrow(1) \,\,\,[\text{(given)}]$

$\,\,\cos^3\theta+\sec^3\theta\\=(\cos\theta+\sec\theta)^3-3\cos\theta\sec\theta(\cos\theta+\sec\theta)\\=(\sqrt3)^3-3\times 1\times \sqrt3\quad[\text{By (1)}]\\=3\sqrt3-3\sqrt3\\=0\,\,\text{(proved)}$

$\,8.\,$ Express $\,\,1+4\csc^2\alpha \cot^2\alpha \,\,$ in the form of a perfect square.

Sol. $\,\,1+4\csc^2\alpha \cot^2\alpha\\=1+4(1+\cot^2\alpha )\cot^2\alpha\,\,\,[**] \\=1+4\cot^2\alpha+4\cot^4\alpha \\=1^2+2\times 1\times 2\cot^2\alpha +(2\cot^2\alpha )^2\\=(1+2\cot^2\alpha )^2\\=(1+\cot^2\alpha +\cot^2\alpha )^2\\=(\csc^2\alpha +\cot^2\alpha )^2 $

Note[**] : $\,\,\csc^2\alpha =1+\cot^2\alpha .$

Here, by $\,\,\csc\alpha \,$ we mean $\text{cosec}\,\alpha. $

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$\,9(i).\,\,$ If $\,3\cos\alpha -4\sin\alpha =5,\,\,$ show that, $\,\,3\sin\alpha +4\cos\alpha =0.$

Sol. We have , $\,3\cos\alpha -4\sin\alpha =5\rightarrow(1)$

So,  $\,\,(3\sin\alpha +4\cos\alpha )^2\\=(3\sin\alpha )^2+2.3\sin\alpha .4\cos\alpha +(4\cos\alpha )^2\\=9\sin^2\alpha +24\sin\alpha \cos\alpha +16\cos^2\alpha \\=9(1-\cos^2\alpha )+24\sin\alpha \cos\alpha \\+16(1-\sin^2\alpha )\\=9+16-(9\cos^2\alpha -24\sin\alpha \cos\alpha \\+16\sin^2\alpha )\\=25-[(3\cos\alpha )^2-2.3\cos\alpha .4\sin\alpha \\+(4\sin\alpha )^2]\\=25-(3\cos\alpha -4\sin\alpha )^2\\=25-5^2\,\,\,[\text{By (1)}]\\=25-25\\=0 \\ \therefore 3\sin\alpha +4\cos\alpha =0\,\,\text{(proved)}$

$\,9(ii)\,\,$ If $\,\,\sin\alpha =x\,\,$ and $\,\,\,\tan\alpha =y,\,\,$ prove that,  $\,\,\frac{1}{x^2}-\frac{1}{y^2}=1.$

Sol. We have, $\,\,x=\sin\alpha  \Rightarrow \frac 1x=\csc\alpha \rightarrow(1) \\ y=\tan\alpha  \Rightarrow \frac 1y=\cot\alpha \rightarrow(2)$

So, $\,\,\frac{1}{x^2}-\frac{1}{y^2}\\=\csc^2\alpha -\cot^2\alpha \,\,\,[\text{By (1), (2)}]\\=1\,\,\,\text{(proved)}$

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

$\,\,10(i)\,\,$ If $\,\,\tan\alpha+\cot\alpha=2,\,\,$ show that, $\,\,\tan^7\alpha+\cot^7\alpha=2.$

Sol.  $\,\,\tan\alpha+\cot\alpha=2\,\,\text{(given)}\rightarrow(1)$

Now, let $\,\,\tan\alpha=x,\,\,\text{so that },\,\,\cot\alpha=\frac 1x.$

Then by $\,(1)\,$ we have , $\,\,x+\frac 1x=2 \\ \Rightarrow x^2+1=2x \\ \Rightarrow x^2-2x+1=0 \\ \Rightarrow (x-1)^2-0 \\ \Rightarrow x-1= 0 \\ \Rightarrow x=1\ \rightarrow(2)$

So, $\,\,\tan^7\alpha+\cot^7\alpha\\=x^7+\frac{1}{x^7}\\=1^7+\frac{1}{1^7}\,\,\,[\text{By (2)}]\\=1+1\\=2\,\,\,\text{(showed)}$

$\,\,10(ii)\,\,$ If $\,\,a_n=\sin^n\theta+\csc^n\theta\,\,$ and $\,\,a_1=2,\,\,$ prove that, $\,\,a_n=2.$

Sol. $\,\,a_n=\sin^n\theta+\csc^n\theta\,\,\text{(given)}\\ \Rightarrow a_1=\sin\theta+\csc\theta \\ \Rightarrow 2=x+\frac 1x\,\,\,[\text{let,}\,\,x=\sin\theta]\\ \Rightarrow 2x=x^2+1 \\ \Rightarrow 0=x^2-2x+1 \\ \Rightarrow (x-1)^2=0 \\ \Rightarrow x-1=0 \\ \Rightarrow x=1 \\ \therefore \sin\theta=1,\,\,\csc\theta=1\rightarrow(1)$

Hence, $\,\,a_n=\sin^n\theta+\csc^n\theta \\ \Rightarrow a_n=1^n+1^n\,\,\,[\text{By (1)}]\\~~~~~~~~~=1+1\\~~~~~~~~~=2\,\,\,\text{(proved)}$

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

$\,\,10(iii)\,\,$ If $\,\,\cos\alpha +\sec\alpha =2,\,\,$ show that, $\,\,\cos^3\alpha +\sec^3\alpha =2.$

Sol. We have, $\,\,\cos\alpha +\sec\alpha =2 \\ \Rightarrow \cos\alpha +\frac{1}{\cos\alpha }=2 \\ \Rightarrow \cos^2\alpha +1=2\cos\alpha  \\ \Rightarrow \cos^2\alpha -2\cos\alpha +1=0 \\ \Rightarrow (\cos\alpha -1)^2=0 \\ \Rightarrow \cos\alpha -1=0 \\ \Rightarrow \cos\alpha =1 \rightarrow(1) \\ \therefore \sec\alpha =\frac{1}{\cos\alpha }=\frac 11=1\rightarrow(2)$

So, $\,\,\cos^3\alpha +\sec^3\alpha \\=1^3+1^3\,\,[\text{by (1), (2)}]\\=1+1\\=2\,\,\,\text{(showed)}$ 

$\,\,11(i)\,\,$ If $\,\,x=r\cos\theta\cos\phi,\,\,y=r\cos\theta\sin\phi, \\z=r\sin\theta,\,\,$ show that, $\,\,x^2+y^2+z^2=r^2.$

Sol.  We have, $\,\,x=r\cos\theta\cos\phi \rightarrow(1)\\y=r\cos\theta\sin\phi \rightarrow(2)\\z=r\sin\theta \rightarrow(3)$

So, $\,\,x^2+y^2+z^2\\=r^2\cos^2\theta\cos^2\phi+r^2\cos^2\theta\sin^2\phi\\+r^2\sin^2\theta\quad[\text{By (1),(2),(3)}]\\=r^2\cos^2\theta(\cos^2\phi+\sin^2\phi)+r^2\sin^2\theta\\=r^2\cos^2\theta\times 1+r^2\sin^2\theta\\=r^2(\cos^2\theta+\sin^2\theta)\\=r^2\times 1\\=r^2\,\,\,\text{(showed)}$ 

$\,\,11(ii)\,\,$ If $\,\,x=a\sec\theta\cos\phi,\,\,y=b\sec\theta\sin\phi,\\z=c\tan\theta,\,\,$ 

show that, $\,\,\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$

Sol. We have, $\,\,x=a\sec\theta\cos\phi\rightarrow(1)\\y=b\sec\theta\sin\phi\rightarrow(2)\\z=c\tan\theta\rightarrow(3)$

Now, $\,\,\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\\=\sec^2\theta\cos^2\phi+\sec^2\theta\sin^2\phi\\-\tan^2\theta\,\,\,[\text{By (1),(2),(3)}]\\=\sec^2\theta(\cos^2\phi+\sin^2\phi)-\tan^2\theta\\=\sec^2\theta\times 1-\tan^2\theta\\=\sec^2\theta-\tan^2\theta\\=1\,\,\,\text{(showed)}$