Differentiation (Part-38) | S N De

 

$\,\,13.\,\,$ Eliminate $\,\theta\,$ in each of the following : 

$\,(i)\,\,x=a\sec\theta,\,\,y=b\tan\theta $

Sol. $\,\,x=a\sec\theta \Rightarrow \sec\theta=\frac xa \rightarrow(1) \\ y=b\tan\theta \Rightarrow \tan\theta=\frac yb \rightarrow(2)$

We know, $\,\,\sec^2\theta-\tan^2\theta=1 \\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\,\,[\text{By (1),(2)}]$

$\,(ii)\,\,a\sin\theta=p,\,\,\,b\tan\theta=q$

Sol. We have, $\,\,a\sin\theta=p \Rightarrow \csc\theta=\frac ap\rightarrow(1)  \\ b\tan\theta=q \Rightarrow \cot\theta=\frac bq\rightarrow(2)$

We know, $\,\,\csc^2\theta-\cot^2\theta=1 \\ \Rightarrow \frac{a^2}{p^2}-\frac{b^2}{q^2}=1\,\,[\text{By (1),(2)}]$

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

$\,\,(iii)\,\,\sin\theta+\cos\theta=m,\,\,\,\tan\theta+\cot\theta=n.$

Sol. We have , $\,\,\tan\theta+\cot\theta=n \\ \Rightarrow \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=n \\ \Rightarrow \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=n \\ \Rightarrow \frac{1}{\sin\theta\cos\theta}=n \\ \Rightarrow \sin\theta\cos\theta=\frac 1n \rightarrow(1)$

We are given, $\,\,\sin\theta+\cos\theta=m \\ \Rightarrow (\sin\theta+\cos\theta)^2=m^2 \\ \Rightarrow \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=m^2 \\ \Rightarrow 1+\frac 2n=m^2\,\,\,\,\,\,[\text{By (1)}] \\ \Rightarrow \frac 2n=m^2-1 \\ \Rightarrow2=n(m^2-1)$

$\,\,(iv)\,\,\sin\theta-\cos\theta=a,\,\,\,\sec\theta-\csc\theta=b $

Sol.  $\,\,\sin\theta-\cos\theta=a \rightarrow(1) \,\,[\text{given}]$

We have, $\,\,\sec\theta-\csc\theta=b \\ \Rightarrow \frac{1}{\cos\theta}-\frac{1}{\sin\theta}=b \\ \Rightarrow \frac{\sin\theta-\cos\theta}{\sin\theta\cos\theta}=b \\ \Rightarrow \frac{a}{\sin\theta\cos\theta}=b\,\,\,\,\,[\text{By (1)}] \\ \Rightarrow \sin\theta\cos\theta=\frac ab\rightarrow(2)$

From $\,\,(1)\,\,$ we get, $\,\,\,(\sin\theta-\cos\theta)^2=a^2 \\ \Rightarrow \sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=a^2 \\ \Rightarrow 1-\frac{2a}{b}=a^2 \,\,\,[\text{By (2)}] \\ \Rightarrow 1-a^2=\frac{2a}{b}\\ \Rightarrow b(1-a^2)=2a$ 

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$\,\,14.\,\,$ If $\,\,7\sin^2\theta+3\cos^2\theta=4,\,\,$ prove that, $\,\,\tan\theta=\pm \frac{1}{\sqrt3}.$

Sol. $\,\,7\sin^2\theta+3\cos^2\theta=4 \\ \Rightarrow 7\sin^2\theta+3(1-\sin^2\theta)=4 \\ \Rightarrow 7\sin^2\theta+3-3\sin^2\theta=4 \\ \Rightarrow 4\sin^2\theta=4-3 \\ \Rightarrow \sin^2\theta=\frac{1}{4}\rightarrow(1) \\ \therefore \cos^2\theta=1-\sin^2\theta\\~~~~~~~~~~~~~=1-\frac 14\\~~~~~~~~~~~~~=\frac 34\rightarrow(2) \\ \therefore \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{\frac 14}{\frac 34}=\frac 13 \\ \Rightarrow \tan\theta=\pm \frac{1}{\sqrt3}\,\,\,\,\text{(proved)}$

$\,\,15.\,\,$ Find the minimum value of $\,\,\cos^4\theta+\sin^2\theta.$

Sol. We have, $\,\,\cos^4\theta+\sin^2\theta\\=\cos^4\theta+1-\cos^2\theta\\=(\cos^2\theta)^2-2.\cos^2\theta.\frac 12+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\\=(\cos^2\theta-\frac 12)^2+\frac 34\rightarrow(1)$

From $\,\,(1),\,\,$ we see the minimum value of $\,\,(\cos^2\theta-\frac 12)^2\,\,$ is $\,\,0.\,\,$ So, the minimum value of $\,\,\cos^4\theta+\sin^2\theta\,\,$ is $\,\,\frac 34\,\,[\text{By (1)}].$