Differentiation (Part-38) | S N De

 

$\,4.\,$ If $\,\,\tan\theta+\sin\theta=m\,\,$ and $\,\tan\theta-\sin\theta=n,\,$ prove that, $\,\,mn=\tan^2\theta\sin^2\theta\,\,$ and $\,\,m^2-n^2=4\sqrt{mn}\,\,(m>n).$

Sol. We have, $\,\,\tan\theta+\sin\theta=m \rightarrow(1)\\ \tan\theta-\sin\theta=n\rightarrow(2) \\ \therefore (\tan\theta+\sin\theta)(\tan\theta-\sin\theta)=mn \\ \Rightarrow \tan^2\theta- \sin^2\theta=mn \\ \Rightarrow \sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right)=mn \\ \Rightarrow \sin^2\theta \times \frac{1-\cos^2\theta}{\cos^2\theta}=mn \\ \Rightarrow \sin^2\theta \times \frac{\sin^2\theta}{\cos^2\theta}=mn \\ \Rightarrow \sin^2\theta\tan^2\theta=mn \rightarrow(3)\,\,\text{(proved)}$

2nd Part :

$\,\,m^2-n^2\\=(m+n)(m-n)\\=2\tan\theta \times 2\sin\theta\,\,\,\,[**]\\=4\tan\theta\sin\theta\\=4\sqrt{mn}\,\,\,\,\,[\text{By (3)}] \\ \text{(proved)}$

Note[**] : Adding $\,(1)\,$ and $\,(2)\,$ we get, 

$m+n=(\tan\theta+\sin\theta)+(\tan\theta-\sin\theta)\\~~~~~~~~~~~=2\tan\theta$

Again, subtracting $\,(2)\,$ from $\,(1)\,$ we get,

$m-n=(\tan\theta+\sin\theta)-(\tan\theta-\sin\theta)\\~~~~~~~~~~~=2\sin\theta$ 

$\,5.\,$ If $\,\csc \alpha-\sin\alpha= m^3 ;\,\,\sec\alpha-\cos\alpha=n^3 ,\,\,$ show that , $m^2n^2(m^2+n^2)=1.$

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

Sol. $\,\csc \alpha-\sin\alpha= m^3 \\ \Rightarrow \frac{1}{\sin\alpha}-\sin\alpha=m^3 \\ \Rightarrow \frac{1-\sin^2\alpha}{\sin\alpha}=m^3 \\ \Rightarrow \frac{\cos^2\alpha}{\sin\alpha}=m^3 \\ \Rightarrow m=\frac{\cos^{\frac23}\alpha}{\sin^{\frac 13}\alpha}\rightarrow(1)$

Again, $\sec\alpha-\cos\alpha=n^3 \\ \Rightarrow \frac{1}{\cos\alpha}-\cos\alpha=n^3 \\ \Rightarrow \frac{1-\cos^2\alpha}{\cos\alpha}=n^3 \\ \Rightarrow \frac{\sin^2\alpha}{\cos\alpha}=n^3 \\ \Rightarrow n=\frac{\sin^{\frac 23}\alpha}{\cos^{\frac 13}\alpha}\rightarrow(2)$

Now, $\,\,m^2n^2\\=\left(\frac{\cos^{\frac23}\alpha}{\sin^{\frac 13}\alpha}\right)^2 \times \left(\frac{\sin^{\frac 23}\alpha}{\cos^{\frac 13}\alpha}\right)^2\\=\frac{\cos^{\frac43}\alpha}{\sin^{\frac 23}\alpha} \times \frac{\sin^{\frac 43}\alpha}{\cos^{\frac 23}\alpha}\\=\cos^{\frac 23}\alpha \times  \sin^{\frac 23}\alpha \rightarrow(3)$

Also, $\,\,m^2+n^2\\=\left(\frac{\cos^{\frac23}\alpha}{\sin^{\frac 13}\alpha}\right)^2 + \left(\frac{\sin^{\frac 23}\alpha}{\cos^{\frac 13}\alpha}\right)^2\\=\frac{\cos^{\frac43}\alpha}{\sin^{\frac 23}\alpha}+\frac{\sin^{\frac 43}\alpha}{\cos^{\frac 23}\alpha}\\=\frac{\cos^2\alpha+\sin^2\alpha}{\sin^{\frac 23}\alpha \times \cos^{\frac 23}\alpha}\\=\frac{1}{\sin^{\frac 23}\alpha \times \cos^{\frac 23}\alpha} \rightarrow(4)$

Multiplying $\,(3)\,$  with $\,(4)\,$ we get, 

$m^2n^2(m^2+n^2)\\=\cos^{\frac 23}\alpha \times  \sin^{\frac 23}\alpha \times \frac{1}{\sin^{\frac 23}\alpha \times \cos^{\frac 23}\alpha}\\=1\,\,\text{(proved)}$

$\,6.\,$ If $\,\,\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y},\,\,$ show that , $\,\,\frac{\sin^{12}\theta}{x^5}+\frac{\cos^{12}\theta}{y^5}=\frac{1}{(x+y)^5}.$

Sol. We have, $\,\,\frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{1}{x+y}\\ \Rightarrow \frac{\sin^4\theta}{x}+\frac{\cos^4\theta}{y}=\frac{\sin^2\theta+\cos^2\theta}{x+y} \\ \Rightarrow \frac{\sin^4\theta}{x}-\frac{\sin^2\theta}{x+y}=\frac{\cos^2\theta}{x+y}-\frac{\cos^4\theta}{y} \\ \Rightarrow \frac{(x+y)\sin^4\theta-x\sin^2\theta}{x(x+y)}=\frac{y\cos^2\theta-(x+y)\cos^4\theta}{y(x+y)} \\ \Rightarrow \frac{y\sin^4\theta-x\sin^2\theta(1-\sin^2\theta)}{x}=\frac{y\cos^2\theta(1-\cos^2\theta)-x\cos^4\theta}{y}\\ \Rightarrow \frac{y\sin^4\theta-x\sin^2\theta\cos^2\theta}{x}=\frac{y\cos^2\theta\sin^2\theta-x\cos^4\theta}{y}\\ \Rightarrow \frac{(y\sin^2\theta-x\cos^2\theta)\sin^2\theta}{x}=\frac{(y\sin^2\theta-x\cos^2\theta)\cos^2\theta}{y}\\ \Rightarrow \frac{\sin^2\theta}{x}=\frac{\cos^2\theta}{y}=\frac{\sin^2\theta+\cos^2\theta}{x+y}=\frac{1}{x+y} \\ \therefore \sin^2\theta=\frac{x}{x+y}, \,\, \cos^2\theta=\frac{y}{x+y}.$

Now, $\,\,\frac{\sin^{12}\theta}{x^5}+\frac{\cos^{12}\theta}{y^5}\\=\frac{1}{x^5}\left(\frac{x}{x+y}\right)^6+\frac{1}{y^5}\left(\frac{y}{x+y}\right)^6\\=\frac{x}{(x+y)^6}+\frac{y}{(x+y)^6}\\=\frac{x+y}{(x+y)^6}\\=\frac{1}{(x+y)^5}\,\,\,\,\text{(showed)}$

$\,7.\,$ If $\,\,\csc\alpha+\cot\alpha=2+\sqrt5 ,\,\,$ show that, $\,\,\cos\alpha=\frac{2}{\sqrt5}.$

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

Sol. We have , $\,\,\csc\alpha+\cot\alpha=2+\sqrt5 \rightarrow(1)$

We know, $\,\,\csc^2\alpha-\cot^2\alpha=1 \\ \Rightarrow (\csc\alpha+\cot\alpha)(\csc\alpha-\cot\alpha)=1 \\ \Rightarrow \csc\alpha-\cot\alpha=\frac{1}{\csc\alpha+\cot\alpha}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2+\sqrt5}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt5-2}{(\sqrt5+2)(\sqrt5-2)}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt5-2}{(\sqrt5)^2-2^2}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt5-2}{5-4}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=\sqrt5-2 \rightarrow(2)$

Adding $\,(1)\,$ and $\,(2),\,$ we get 

$\,\,(\csc\alpha+\cot\alpha)+(\csc\alpha-\cot\alpha)\\=2+\sqrt5+\sqrt5-2 \\ \Rightarrow 2\csc\alpha=2\sqrt5 \\ \Rightarrow \csc\alpha=\sqrt5 \\ \Rightarrow \sin\alpha=\frac{1}{\sqrt5}\rightarrow(3)$

Hence, $\,\,\cos\alpha\\=\sqrt{1-\sin^2\alpha}\\=\sqrt{1-\left(\frac{1}{\sqrt5}\right)^2}\,\,[\text{By (3)}]\\=\sqrt{1-\frac 15}\\=\sqrt{\frac 45}\\=\frac{2}{\sqrt5}.$

$\,8.\,$ If $\,\,\tan A =n\tan B,\,\,\sin A=m\sin B,\,\,$ prove that, $\,\,\cos^2A=\frac{m^2-1}{n^2-1}.$

Sol. $\,\,\tan A =n\tan B \\ \Rightarrow \tan B=\frac{\tan A}{n} \\ \Rightarrow \cot B=\frac{n}{\tan A}\rightarrow(1)$

Again, $\,\sin A=m\sin B \\ \Rightarrow \sin B=\frac{\sin A}{m}\\ \Rightarrow \csc B=\frac{m}{\sin A}\rightarrow(2)$

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

Now, $\,\,\csc^2B-\cot^2B=1 \\ \Rightarrow \frac{m^2}{\sin^2A}-\frac{n^2}{\tan^2A}=1 \\ \Rightarrow \frac{m^2}{\sin^2A}-\frac{n^2\cos^2A}{\sin^2A}=1 \\ \Rightarrow \frac{m^2-n^2\cos^2A}{\sin^2A}=1\\ \Rightarrow m^2-n^2\cos^2A=\sin^2A \\ \Rightarrow m^2-n^2\cos^2A=1-\cos^2A \\ \Rightarrow m^2-1=(n^2-1)\cos^2A \\ \Rightarrow \cos^2A=\frac{m^2-1}{n^2-1}\,\,\text{(proved)}$