Differentiation (Part-38) | S N De

 $\,2(i).\,$ If $\,\tan^2\alpha=1+2\tan^2\beta,\,\,$ prove that $\,\,\cos 2\beta=1+2\cos2\alpha$

Sol. $\,\tan^2\alpha=1+2\tan^2\beta \\ \Rightarrow \tan^2\beta=\frac 12(\tan^2\alpha-1) \\ \therefore \,\cos2\beta=\frac{1-\tan^2\beta}{1+\tan^2\beta}\\~~~~~~~~~~~~~~=\frac{1-\frac 12(\tan^2\alpha-1)}{1+\frac 12(\tan^2\alpha-1)}\\~~~~~~~~~~~~~~=\frac{3-\tan^2\alpha}{1+\tan^2\alpha}\\~~~~~~~~~~~~~~=\frac{1+\tan^2\alpha+2(1-\tan^2\alpha)}{1+\tan^2\alpha}\\~~~~~~~~~~~~~~=1+\frac{2(1-\tan^2\alpha)}{1+\tan^2\alpha}\\~~~~~~~~~~~~~~=1+2\cos2\alpha$

$\,2(ii)\,$ If $\,\,\sin\alpha+\cos\alpha=\sqrt2\cos\alpha,\,\,$ show that, $\,\tan2\alpha=1.$

Sol.  $\,\,\sin\alpha+\cos\alpha=\sqrt2\cos\alpha\\ \Rightarrow \tan\alpha+1=\sqrt2 \\ \Rightarrow \tan\alpha=\sqrt2-1 \\ \text{Now,}\,\,\,\tan2\alpha\\=\frac{2\tan\alpha}{1-\tan^2\alpha}\\=\frac{2(\sqrt2-1)}{1-(\sqrt2-1)^2}\\=\frac{2(\sqrt2-1)}{1-2+2\sqrt2-1}\\=\frac{2(\sqrt2-1)}{2(\sqrt2-1)}\\=1$

$\,3.\,$ If $\,\,\tan\theta=\frac 17,\,\,\tan \phi=\frac 13,\,\,$ show that, $\,\cos 2\theta=\sin4\phi$

Sol. $\,\,\cos2\theta =\frac{1-\tan^2\theta}{1+\tan^2\theta}\\~~~~~~~~~~~=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}\\~~~~~~~~~~~=\frac{48}{50}\\~~~~~~~~~~~=\frac{24}{25}\cdots(1)$  

Again, $\,\,\sin4\phi=\frac{2\tan2\phi}{1+\tan^22\phi}\\~~~~~~~~~~~=\frac{2 \times \frac{2\tan\phi}{1-\tan^2\phi}}{1+\left(\frac{2\tan\phi}{1-\tan^2\phi}\right)^2}\\~~~~~~~~~~~=\frac{4\tan\phi(1-\tan^2\phi)}{(1-\tan^2\phi)^2+4\tan^2\phi}\\~~~~~~~~~~~=\frac{4\times \frac 13\left(1-\frac 19\right)}{\left(1-\frac 19\right)^2+4\times \frac 19}\\~~~~~~~~~~~=\frac{4\times \frac 13\times\frac 89 \times9^2}{8^2+4 \times9}\\~~~~~~~~~~~=\frac{24}{25}\cdots(2) \\~~~~~~~~~~~ \therefore \cos2\theta=\sin4\phi.\,\,[\text{By (1) and (2)}]$

$\,4.\,$ If $\,\,\tan\theta=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha},\,\,$prove that, $\,\,2\cos^2\theta=1+\sin2\alpha$

Sol. $\,\,2\cos^2\theta=\frac{2}{\sec^2\theta}\\~~~~~~~~~~~~~~=\frac{2}{1+\tan^2\theta}\\~~~~~~~~~~~~~~=\frac{2}{1+\left(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\right)^2}\\~~~~~~~~~~~~~~=\frac{2(\sin\alpha+\cos\alpha)^2}{(\sin\alpha+\cos\alpha)^2+(\sin\alpha-\cos\alpha)^2}\\~~~~~~~~~~~~~~=\frac{2(\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha)}{2(\sin^2\alpha+\cos^2\alpha)}\\~~~~~~~~~~~~~~=\frac{1+\sin2\alpha}{1}\\~~~~~~~~~~~~~~=1+\sin2\alpha$

$\,5.\,$ If $\,\,\tan^2\theta=1+2\tan^2\phi,\,\,$show that, $\,\cos2\theta+\sin^2\phi=0$

Sol. $\,\cos2\theta+\sin^2\phi\\=\frac{1-\tan^2\theta}{1+\tan^2\theta}+\sin^2\phi\\=\frac{1-1-2\tan^2\phi}{1+1+2\tan^2\phi}+\sin^2\phi\\=\frac{-2\tan^2\phi}{2(1+\tan^2\phi)}+\sin^2\phi\\=-\frac{\tan^2\phi}{\sec^2\phi}+\sin^2\phi\\=-\sin^2\phi+\sin^2\phi\\=0$

$\,6.\,$ If $\,\tan x=\frac ba,\,\,$ find the value of $\,\,a^2\csc 2x+b^2\sec2x.$

Sol. $\,\,a^2\csc 2x+b^2\sec2x\\=\frac{a^2}{\sin2x}+\frac{b^2}{\cos2x}\\=\frac{a^2(1+\tan^2x)}{2\tan x}+\frac{b^2(1+\tan^2x)}{1-\tan^2x}\\=\frac{a^2\left(1+\frac{b^2}{a^2}\right)}{\frac{2b}{a}}+\frac{b^2\left(1+\frac{b^2}{a^2}\right)}{1-\frac{b^2}{a^2}}\\=\frac{a(a^2+b^2)}{2b}+\frac{b^2(a^2+b^2)}{a^2-b^2}\\=(a^2+b^2)\left(\frac{a}{2b}+\frac{b^2}{a^2-b^2}\right)\\=\frac{(a^2+b^2)(a^3-ab^2+2b^3)}{2b(a^2-b^2)}$

Note: By csc , we mean cosec .

$\,7.\,$ If $\,\,2\tan\alpha=3\tan\beta,\,$ show that, $\,\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$

Sol. $\,\,\tan(\alpha-\beta)\\=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\\=\frac{\frac 32\tan\beta-\tan\beta}{1+\frac 32\tan^2\beta}\,\,[\text{Using,}\,2\tan\alpha=3\tan\beta]\\=\frac{\tan\beta}{2+3\tan^2\beta}\\=\frac{2\tan\beta}{4+6\tan^2\beta}\\=\frac{\frac{2\tan\beta}{1+\tan^2\beta}}{\frac{5(1+\tan^2\beta)-(1-\tan^2\beta)}{1+\tan^2\beta}}\\=\frac{\sin2\beta}{5-\frac{1-\tan^2\beta}{1+\tan^2\beta}}\\=\frac{\sin2\beta}{5-\cos2\beta}$

$\,8.\,$ If $\,\frac{\pi}{2}<\theta<\pi\,\,$ and $\,5\sin^2\theta+3\cos^2\theta=4,\,\,$ find the values of $\,\,\sin2\theta\,\,$ and $\,\cos3\theta.$

Sol. $\,5\sin^2\theta+3\cos^2\theta=4 \\ \Rightarrow 5\sin^2\theta+3-3\sin^2\theta=4 \\ \Rightarrow \sin^2\theta=\frac 12 \\ \Rightarrow \sin\theta=\frac{1}{\sqrt2}\\ [\text{Since,}\,\,\frac{\pi}{2}<\theta<\pi,\,\,\,\,\,\therefore \,\,\sin\theta>0] \\ \Rightarrow \theta=\frac{3\pi}{4}\,\,\,\,[\frac{\pi}{2}<\theta<\pi]\\ \text{Now,}\,\,\sin2\theta=\sin(2\times \frac{3\pi}{4})=\sin \frac{3\pi}{2}=-1 \\ \text{and}\,\,\,\cos3\theta\\=\cos\frac{9\pi}{4}\\=\cos(2\pi+\frac{\pi}{4})\\=\cos\frac{\pi}{4}\\=\frac{1}{\sqrt2}$

$\,9.\,$ If $\,\,x^2+y^2=1,\,$ show that, $\,(3x-4x^3)^2+(3y-4y^3)^2=1$

Sol. Since $\,\,x^2+y^2=1 \cdots(1),\,\text{so,}\,\,-1\leq x\leq1, \\ \text{and}\,\,\,\,\,\,-1\leq y\leq 1.$ 

Let $\,\,x=\sin\theta,\,\,y=\cos\theta\,\,$ which satisfy (1). 

Now,  $\,(3x-4x^3)^2+(3y-4y^3)^2\\=(3\sin\theta-4\sin^3\theta)^2+(3\cos\theta-4\cos^3\theta)^2\\=(\sin3\theta)^2+(-\cos3\theta)^2\\=\sin^23\theta+\cos^23\theta\\=1$