Differentiation (Part-38) | S N De

 $\,10(i).\,\,$ If $\,\sin^4\theta+\sin^2\theta=1,\,\,$ show that , $\,\tan^4\theta-\tan^2\theta=1.$

Sol. We have ,  $\,\sin^4\theta+\sin^2\theta=1\\ \Rightarrow \sin^4\theta=1-\sin^2\theta\\~~~~~~~~~~~~~~=\cos^2\theta\rightarrow(1)$

Now, L.H.S$\,=\,\tan^4\theta-\tan^2\theta\\=\frac{\sin^4\theta}{\cos^4\theta}-\tan^2\theta\\=\frac{\cos^2\theta}{\cos^4\theta}-\tan^2\theta\,\,[\text{By (1)}]\\=\frac{1}{\cos^2\theta}-\tan^2\theta\\=\sec^2\theta-\tan^2\theta\\=1\\=\text{R.H.S.(proved)}$ 

$\,10(ii)\,\,$ If $\,\,\tan^4\theta+\tan^2\theta=1,\,\,$prove that, $\,\,\cos^4\theta+\cos^2\theta=1.$

Sol. $\,\,\tan^4\theta+\tan^2\theta=1 \\ \Rightarrow \frac{\sin^4\theta}{\cos^4\theta}+\frac{\sin^2\theta}{\cos^2\theta}=1 \\ \Rightarrow \frac{\sin^4\theta+\sin^2\theta\cos^2\theta}{\cos^4\theta}=1 \\ \Rightarrow \frac{\sin^2\theta(\sin^2\theta+\cos^2\theta)}{\cos^4\theta}=1 \\ \Rightarrow \frac{\sin^2\theta}{\cos^4\theta}=1 \\ \Rightarrow \sin^2\theta=\cos^4\theta\rightarrow(1)$

Now, L.H.S. $\,=\cos^4\theta+\cos^2\theta\\=\sin^2\theta+\cos^2\theta\,\,\,\,\,[\text{By (1)}]\\=1=\text{R.H.S.(proved)}$

$\,10(iii)\,\,$ If $\,\,\sin^2\theta=\cos^3\theta,\,\,$ prove that, $\,\,\cot^6\theta-\cot^2\theta=1.$

Sol. We have, $\,\,\sin^2\theta=\cos^3\theta \\ \Rightarrow \frac{\sin^2\theta}{\sin^3\theta}=\frac{\cos^3\theta}{\sin^3\theta} \\ \Rightarrow \frac{1}{\sin\theta}=\cot^3\theta \\ \Rightarrow \csc\theta=\cot^3\theta\rightarrow(1)$

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

Now, L.H.S. $=\cot^6\theta-\cot^2\theta\\=(\cot^3\theta)^2-\cot^2\theta\\=\csc^2\theta-\cot^2\theta\,\,\,[\text{By (1)}]\\=1=\text{R.H.S. (proved)}$

$\,10(iv)\,\,$ If $\,\,\sin\theta+\sin^2\theta=1,\,\,$ show that, $\,\,\cos^4\theta+\cos^2\theta=1.$ 

Sol. $\,\,\sin\theta+\sin^2\theta=1,\,\,[\text{(Given)}] \\ \Rightarrow \sin\theta=1-\sin^2\theta\\~~~~~~~~~~~~=\cos^2\theta\rightarrow(1)$

Now, L.H.S.$\,=\cos^4\theta+\cos^2\theta\\=(\cos^2\theta)^2+\cos^2\theta\\=\sin^2\theta+\cos^2\theta\,\,\,[\text{By (1)}]\\=1=\text{R.H.S.(proved)}$

$\,11.\,$ If $\,\,\cos\theta-\sin\theta=\sqrt2\sin\theta,\,\,$ show that, $\,\,\cos\theta+\sin\theta=\sqrt2\cos\theta.\,\,[0<\theta<\frac{\pi}{2}]$

Sol. We have, $\,\,\cos\theta-\sin\theta=\sqrt2\sin\theta\\ \Rightarrow \cos\theta=(\sqrt2+1)\sin\theta \\ \Rightarrow (\sqrt2-1)\cos\theta=(\sqrt2-1)(\sqrt2+1)\sin\theta \\ \Rightarrow (\sqrt2-1)\cos\theta=[(\sqrt2)^2-1^2]\sin\theta \\ \Rightarrow (\sqrt2-1)\cos\theta=\sin\theta \\ \Rightarrow \sqrt2\cos\theta=\sin\theta+\cos\theta\,\,\,\text{(proved)}$

$\,\,12.\,\,$ If $\,\,u_n=\sin^n\alpha+\cos^n\alpha,\,\,$then show that , the value of $\,\,6u_4-4u_6\,\,$ is independent of $\,\,\alpha.$

Sol.  Let $\,\,\sin\alpha=x,\,\,\cos\alpha=y,\,\,$ so that $\,\,x^2+y^2=1\rightarrow(1)$ and $\,\,u_n=x^n+y^n.$

Now, $\,\,6u_4-4u_6\\=6(x^4+y^4)-4(x^6+y^6)\\=6[(x^2)^2+(y^2)^2]-4[(x^2)^3+(y^2)^3]\\=6[(x^2+y^2)^2-2x^2y^2]\\-4[(x^2+y^2)^3-3x^2y^2(x^2+y^2)]\\=6[1-2x^2y^2]-4[1-3x^2y^2\times1]\,\,[\text{By (1)}]\\=6-12x^2y^2-4+12x^2y^2\\=2 \rightarrow(2)$

From (2), it follows that the value of $\,\,6u_4-4u_6\,\,$ is independent of $\,\,\alpha.$

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