Differentiation (Part-38) | S N De

 $\,2.\,$ If $\,\,n\tan \alpha=(n+1)\tan \beta,\rightarrow(1)\,\,$then show that $\,\,\tan(\alpha-\beta)=\frac{\sin 2 \beta}{2n+1-\cos 2\beta}$ 

Sol. $\,\tan(\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha\tan \beta}\\ \Rightarrow \tan(\alpha-\beta)=\frac{\frac{n+1}{n}\tan \beta-\tan \beta}{1+\frac{n+1}{n}\tan^2\beta}\,\,[\text{By (1)}]\\ \Rightarrow \tan(\alpha-\beta)=\frac{\sin \beta}{n \cos \beta} \frac{n \cos^2\beta}{(n \cos^2\beta+n \sin^2 \beta+\sin^2\beta)}\\ \Rightarrow \tan(\alpha-\beta)=\frac{\sin \beta\cos\beta}{n+\sin^2\beta}\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\sin 2\beta}{2n+2\sin^2\beta}\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{\sin 2\beta}{2n+1-\cos 2\beta}$

$\,3.\,$ If $\,\,\tan \theta=\sec 2\alpha,\,\,$ show that $$\sin 2\theta=\frac{1-\tan^4\alpha}{1+\tan^4\alpha}$$

Sol.  $\,\,\sin 2\theta=\frac{2\tan \theta}{1+\tan^2\theta}\\~~~~~~~~~~~=\frac{2\sec 2\alpha}{1+\sec^22\alpha} \\~~~~~~~~~~~=\frac{2.\frac{1+\tan^2\alpha}{1-\tan^2\alpha}}{1+\left(\frac{1+\tan^2\alpha}{1-\tan^2\alpha}\right)^2}\\~~~~~~~~~~~=\frac{2(1+\tan^2\alpha)}{1-\tan^2\alpha}.\frac{(1-\tan^2\alpha)^2}{(1-\tan^2\alpha)^2+(1+\tan^2\alpha)^2}\\~~~~~~~~~~~=\frac{2(1-\tan^4\alpha)}{2(1+\tan^4\alpha)}\\ \therefore \sin 2\theta=\frac{1-\tan^4\alpha}{1+\tan^4\alpha}$

$\,4.\,$ Prove that $\,\,(i)\,8\cos^4\theta=3+4\cos2\theta+\cos4\theta \\ (ii)\,8\sin^4\theta=3-4\cos2\theta+\cos4\theta\\ (iii)\,64\cos^3\theta\sin^4\theta=\cos 7\theta-\cos5\theta \\-3\cos3\theta+3\cos\theta$

Sol. $\,(i)\, 8\cos^4\theta=2(2\cos^2\theta)^2\\~~~~~~~~~~~~~~~~~~=2(1+\cos 2\theta)^2\\~~~~~~~~~~~~~~~~~~=2(1+2\cos2\theta+\cos^22\theta)\\~~~~~~~~~~~~~~~~~~=2(1+2\cos 2\theta)+2\cos^22\theta\\~~~~~~~~~~~~~~~~~~=2+4\cos 2\theta+1+\cos 4\theta \\~~~~~~~~~~~~~~~~~~=3+4\cos 2\theta+\cos 4\theta$

Sol. $\,(ii)\, 8\sin^4\theta=2(2\sin^2\theta)^2\\~~~~~~~~~~~~~~~~~~~=2(1-\cos 2\theta)^2\\~~~~~~~~~~~~~~~~~~~=2(1-2\cos 2\theta+\cos^22\theta)\\~~~~~~~~~~~~~~~~~~~=2-4\cos2\theta+2\cos^22\theta\\~~~~~~~~~~~~~~~~~~~=2-4\cos 2\theta+1+\cos(2 \times 2\theta)\\~~~~~~~~~~~~~~~~~~=3-4\cos 2\theta+\cos4\theta$

Sol. $\,(iii)\, 64\cos^3\theta\sin^4\theta \\=4.4\cos^3\theta.4\sin^2\theta\\=4.(\cos3\theta+3\cos\theta)(1-\cos2\theta)^2\\=4(\cos3\theta+3\cos\theta)(1-2\cos2\theta+\frac{1+\cos4\theta}{2})\\=2(\cos3\theta+3\cos\theta)(2-4\cos2\theta+1+\cos4\theta)\\=2(\cos3\theta+3\cos\theta)(3-4\cos2\theta+\cos4\theta)\\=2(3\cos 3\theta-4\cos2\theta\cos3\theta+\cos3\theta\cos4\theta\\+9\cos\theta-12\cos\theta\cos2\theta+3\cos\theta\cos4\theta)\\=\cos7\theta-\cos5\theta-3\cos 3\theta+3\cos\theta$

$4(iv)\,$ Prove that $\,\,\sin^2A\cos^4A=\frac{1}{16}\cos2A+\frac{1}{32}\cos 2A\\-\frac{1}{16}\cos4A-\frac{1}{32}\cos6A$

Sol. $\,\,\sin^2A\cos^4A\\=\left(\frac{1-\cos2A}{2}\right) \left(\frac{1+\cos2A}{2}\right)^2\\=\frac 18(1-\cos2A)(1+2\cos2A+\frac{1+\cos4A}{2})\\=\frac{1}{16}(1-\cos2A)(3+4\cos2A+\cos4A)\\=\frac{1}{16}(3+4\cos2A+\cos4A-3\cos2A\\-4\cos^22A-\cos2A\cos4A)\\=\frac{1}{16}(3+\cos2A+\cos4A-2[1+\cos4A] \\-\frac 12\cos6A-\frac 12\cos2A)\\=\frac{1}{16}\left(1+\frac 12\cos2A-\cos4A-\frac 12\cos6A\right)\\=\frac {1}{16}+\frac{1}{32}\cos2A-\frac{1}{16}\cos4A-\frac{1}{32}\cos6A$

$\,5.\,$ If $\,\tan \theta=\frac{\tan \alpha-\tan \beta}{1-\tan \alpha\tan \beta},\,\,$ show that, $\,\,\sin2\theta=\frac{1-\tan^4\alpha}{1+\tan^4\alpha}$

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