Differentiation (Part-38) | S N De

 

$\,11.\,$ If  $\,x-\frac 1x=2i\sin\theta,\,$ show that $\,\,x^4-\frac{1}{x^4}=2i\sin4\theta$

Sol. $\,\,x^4-\frac{1}{x^4}\\=(x^2)^2-\left(\frac{1}{x^2}\right)^2\\=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})\\=\left[(x-\frac 1x)^2+2.x.\frac 1x\right](x+\frac 1x)(x-\frac 1x)\cdots(1)$

$(x+\frac 1x)^2=(x-\frac 1x)^2+4.x.\frac 1x \\ \Rightarrow (x+\frac 1x)=\sqrt{(2i\sin\theta)^2+4}\\~~~~~~~~~~~~~~~~~~~=2\sqrt{i^2\sin^2\theta+1}\\~~~~~~~~~~~~~~~~~~~=2\sqrt{1-\sin^2\theta}\\~~~~~~~~~~~~~~~~~~~=2\cos\theta \cdots(2)$

Hence, from (1) and (2), we get  

$x^4-\frac{1}{x^4}\\=[(2i\sin\theta)^2+2]. 2\cos\theta.2i\sin\theta\\=[2-4\sin^2\theta] \times 4i\sin\theta\cos\theta\\=[2-2(1-\cos2\theta)].2i.\sin2\theta\\=2\cos2\theta.2i\sin2\theta\\=2i(2\sin2\theta\cos2\theta)\\=2i\sin4\theta$

$\,12.\,$  Show that, $\, (x\tan\alpha+y\cot\alpha)(x\cot\alpha+y\tan\alpha)\\=(x+y)^2+4xy\cot^22\alpha$

Sol. We have, $\, (x\tan\alpha+y\cot\alpha)(x\cot\alpha+y\tan\alpha)\\=x^2+y^2+xy\left[\frac{\sin^4\alpha+\cos^4\alpha}{\cos^2\alpha\sin^2\alpha}\right]\\=x^2+y^2+xy\left[\frac{(\sin^2\alpha-\cos^2\alpha)^2+2\sin^2\alpha\cos^2\alpha}{\cos^2\alpha\sin^2\alpha}\right]\\=x^2+y^2+xy\left[\frac{(\cos2\alpha)^2+2\sin^2\alpha\cos^2\alpha}{\cos^2\alpha\sin^2\alpha}\right]\\=x^2+y^2+4xy\left[\frac{\cos^22\alpha+2\sin^2\alpha\cos^2}{4\sin^2\alpha\cos^2\alpha}\right]\\=x^2+y^2+4xy\left[\frac{\cos^22\alpha}{\sin^22\alpha}+\frac 12\right]\\=x^2+y^2+4xy[\cot^2\alpha+\frac 12]\\=x^2+y^2+2xy+4xy\cot^22\alpha\\=(x+y)^2+4xy\cot^22\alpha$

$\,13.\,$ If $\,a\sin\theta+b\cos\theta=c\,\,\text{and}\,\,a\csc\theta+b\sec\theta=c,\,\,$ show that $\,\,\sin 2\theta=\frac{2ab}{c^2-a^2-b^2}.$

Sol. Here, by csc we mean cosec.

Now,$\,\,(a\sin\theta+b\cos\theta)(a\csc\theta+b\sec\theta)=c.c \\ \Rightarrow a^2+ab\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)+b^2=c^2\\ \Rightarrow ab\left(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right)=c^2-a^2-b^2 \\ \Rightarrow ab \left(\frac{1}{\sin\theta\cos\theta}\right)=c^2-a^2-b^2 \\\Rightarrow \frac{2ab}{\sin2\theta}=c^2-a^2-b^2 \\ \Rightarrow \sin2\theta=\frac{2ab}{c^2-a^2-b^2}$

$\,14.\,$ If $\,0 <\alpha<\frac{\pi}{2}\,\,$ and $\,\,\sin \alpha +\cos \alpha+\tan \alpha+\cot \alpha+\sec \alpha\\+\csc \alpha=7\,\,$ 

then show that $\,\,\sin2\alpha\,$ is a root of $\,\,x^2-44x-36=0.$

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Sol. $\,\,\sin \alpha +\cos \alpha+\tan \alpha+\cot \alpha+\sec \alpha+\csc \alpha=7\\ \Rightarrow (\sin \alpha +\cos \alpha)+(\tan \alpha+\cot \alpha)+(\sec \alpha+\csc \alpha)=7 \\ \Rightarrow (\sin \alpha+\cos \alpha)+(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha})+(\frac{1}{\cos \alpha}+\frac{1}{\sin \alpha})=7 \\ \Rightarrow (\sin \alpha+\cos \alpha)+(\frac{\sin^2\alpha+\cos^2\alpha}{\sin \alpha\cos\alpha})+\frac{\sin\alpha+\cos\alpha}{\sin\alpha\cos\alpha}=7 \\ \Rightarrow (\sin \alpha+\cos\alpha)(1+\frac{2}{2\sin\alpha\cos\alpha})+\frac{1}{\sin\alpha\cos\alpha}=7 \\ \Rightarrow (\sin\alpha+\cos\alpha)(1+\frac{2}{\sin2\alpha})=7-\frac{2}{\sin2\alpha} \\ \Rightarrow (1+\sin2\alpha)(1+\frac{4}{\sin2\alpha}+\frac{4}{\sin^22\alpha})=49+\frac{4}{\sin^22\alpha}-\frac{28}{\sin2\alpha}\,\,[**] \\ \Rightarrow (1+\sin2\alpha)(\sin^22\alpha+4\sin2\alpha+4)=49\sin^22\alpha-28\sin2\alpha+4 \\ \Rightarrow \sin^22\alpha-44\sin2\alpha+36=0\,\,[\text{Since,}\,\,\sin2\alpha \neq 0] \cdots(1) \\ [**]=> \text{squaring both sides}$

Hence, by (1) we can say that $\,\,\sin2\alpha\,$ is a root of $\,\,x^2-44x-36=0.$

By  csc we mean cosec.

$\,15.\,$ If $\,\sin(x+y)=\frac{p}{\sqrt{1+p^2}}\,\,$ and $\,\,\cos(x-y)=\frac{1}{\sqrt{1+q^2}},\,$ then show that $\,\,\tan x\,$ is a root of the equation $\,\,(p+q)z^2+2(1-pq)z-(p+q)=0.$

Sol. $\,\sin(x+y)=\frac{p}{\sqrt{1+p^2}}\\ \Rightarrow \cos^2(x+y)=1-\sin^2(x+y)\\~~~~~~~~~~~~~~~~~~~~~~~~=1-\frac{p^2}{1+p^2}\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{1+p^2}\\ \Rightarrow \cos(x+y)=\frac{1}{\sqrt{1+p^2}}$

Again, $\,\,\cos(x-y)=\frac{1}{\sqrt{1+q^2}} \\ \Rightarrow \sin^2(x-y)=1-\cos^2(x-y)\\~~~~~~~~~~~~~~~~~~~~~~~~=1-\frac{1}{1+q^2}\\~~~~~~~~~~~~~~~~~~~~~~~=\frac{q^2}{1+q^2} \\ \Rightarrow \sin(x-y)=\frac{q}{\sqrt{1+q^2}}\\ \sin(2x)=\sin\{(x+y)+(x-y)\}\\~~~~~~~~~~~~~=\sin(x+y)\cos(x-y)+\cos(x+y)\sin(x-y)\\ \Rightarrow \sin(2x)=\frac{p+q}{\sqrt{(1+p^2)(1+q^2)}} \\ \text{Similarly,}\,\,\cos(2x)=\frac{1-pq}{\sqrt{(1+p^2)(1+q^2)}}\\ \therefore \,\, \tan(2x)=\frac{p+q}{1-pq} \\ \Rightarrow \frac{2\tan x}{1-\tan^2x}=\frac{p+q}{1-pq} \\ \Rightarrow (p+q)\tan^2x+2(1-pq)\tan x-(p+q)=0$

So,$\,\,\tan x\,$ is a root of the equation $(p+q)z^2+2(1-pq)z-(p+q)=0.$