Differentiation (Part-38) | S N De

 

$\,\,1.\,\,$ Prove the following identities : 

$\,(i)\,\,\sin A\cos A(\tan A+\cot A)=1$

Sol. $\,\,\sin A \cos A(\tan A+\cot A)\\=\sin A \cos A\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)\\=\sin A\cos A\left(\frac{\sin^2A+\cos^2A}{\sin A\cos A}\right)\\=\sin A\cos A \times \frac{1}{\sin A\cos A}\\=1\quad\text{(proved)}$ 

$\,\,(ii)\,\,\sin\alpha\cos\alpha(\tan\alpha-\cot\alpha)=2\sin^2\alpha-1$

Sol. $\,\,\sin\alpha\cos\alpha(\tan\alpha-\cot\alpha)\\=\sin\alpha\cos\alpha\times \left(\frac{\sin\alpha}{\cos\alpha}-\frac{\cos\alpha}{\sin\alpha}\right)\\=\sin\alpha\cos\alpha \times \frac{\sin^2\alpha-\cos^2\alpha}{\sin\alpha\cos\alpha}\\=\sin^2\alpha-\cos^2\alpha\\=\sin^2\alpha-(1-\sin^2\alpha)\\=2\sin^2\alpha-1\quad\text{(proved)}$

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$\,\,(iii)\,\,\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\sec\theta+\tan\theta$

Sol. $\quad \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\\=\sqrt{\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}}\\=\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}\\=\sqrt{\frac{(1+\sin\theta)^2}{\cos^2\theta}}\\=\frac{1+\sin\theta}{\cos\theta}\\=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\\=\sec\theta+\tan\theta\quad\text{(proved)}$

$\,\,(iv)\,\,\sqrt{\frac{\csc\theta-1}{\csc\theta+1}}=\frac{1-\sin\theta}{\cos\theta}$

Sol. $\quad \sqrt{\frac{\csc\theta-1}{\csc\theta+1}}\\=\sqrt{\frac{\sin\theta(\csc\theta-1)}{\sin\theta(\csc\theta+1)}}\\=\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}\\=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}\\=\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}\\=\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}\\=\frac{1-\sin\theta}{\cos\theta}\quad\text{(proved)}$

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

$\,\,(v)\,\, \sin^4\alpha-\cos^4\alpha+1=2\sin^2\alpha$

Sol. $\quad \sin^4\alpha-\cos^4\alpha+1\\=(\sin^2\alpha)^2-(\cos^2\alpha)^2+1\\=(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)+1\\=1\times (\sin^2\alpha-\cos^2\alpha)+(\sin^2\alpha+\cos^2\alpha)\\=2\sin^2\alpha\quad \text{(proved)}$

$\,\,(vi)\,\,\frac{\cos\phi}{1+\sin\phi}+\frac{1+\sin\phi}{\cos\phi}=2\sec\phi$

Sol. $\quad \frac{\cos\phi}{1+\sin\phi}+\frac{1+\sin\phi}{\cos\phi}\\=\frac{\cos^2\phi+(1+\sin\phi)^2}{\cos\phi(1+\sin\phi)}\\=\frac{\cos^2\phi+1+2\sin\phi+\sin^2\phi}{\cos\phi(1+\sin\phi)}\\=\frac{(\sin^2\phi+\cos^2\phi)+1+2\sin\phi}{\cos\phi(1+\sin\phi)}\\=\frac{1+1+2\sin\phi}{\cos\phi(1+\sin\phi)}\\=\frac{2(1+\sin\phi)}{\cos\phi(1+\sin\phi)}\\=\frac{2}{\cos\phi}\\=2\sec\phi\quad\text{(proved)}$

$\,\,(vii)\,\,\,(\csc\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2\\-(\tan\theta-\cot\theta)^2=1$

Sol. $\quad (\csc\theta-\sin\theta)^2+(\sec\theta-\cos\theta)^2\\-(\tan\theta-\cot\theta)^2\\=(\csc^2\theta-2\csc\theta\sin\theta+\sin^2\theta)\\+(\sec^2\theta-2\sec\theta\cos\theta+\cos^2\theta)\\-(\tan^2\theta-2\tan\theta\cot\theta+\cot^2\theta)\\=\csc^2\theta-2+\sin^2\theta+\sec^2\theta-2\\+\cos^2\theta-\tan^2\theta+2-\cot^2\theta\\=(\csc^2\theta-\cot^2\theta)+(\sec^2\theta-\tan^2\theta)\\+(\sin^2\theta+\cos^2\theta)-2-2+2\\=1+1+1-2-2+2\\=1\quad\text{(proved)}$

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

$\,\,(viii)\,\,(\cot\theta+\csc\theta)^2=\frac{1+\cos\theta}{1-\cos\theta}$

Sol. $\,\,\text{R.H.S.}=\frac{1+\cos\theta}{1-\cos\theta}\\=\frac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\=\frac{(1+\cos\theta)^2}{1-\cos^2\theta}\\=\frac{(1+\cos\theta)^2}{\sin^2\theta}\\=\left(\frac{1+\cos\theta}{\sin\theta}\right)^2\\=\left(\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}\right)^2\\=(\csc\theta+\cot\theta)^2\\=\text{L.H.S. (proved)}$

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

$\,\,(ix)\,\,\frac{1+\sin\theta}{1-\sin\theta}=(\sec\theta+\tan\theta)^2$

Sol.  $\quad \frac{1+\sin\theta}{1-\sin\theta}\\=\frac{(1+\sin\theta)(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\\=\frac{(1+\sin\theta)^2}{1-\sin^2\theta}\\=\frac{(1+\sin\theta)^2}{\cos^2\theta}\\=\left(\frac{1+\sin\theta}{\cos\theta}\right)^2\\=\left(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\right)^2\\=(\sec\theta+\tan\theta)^2\quad\text{(proved)}$

$\,\,(x)\,\,(\sin\alpha+\csc\alpha)^2+(\cos\alpha+\sec\alpha)^2\\=\tan^2\alpha+\cot^2\alpha+7$

Sol. $\quad (\sin\alpha+\csc\alpha)^2+(\cos\alpha+\sec\alpha)^2\\=(\sin^2\alpha+2\sin\alpha\csc\alpha+\csc^2\alpha)\\+(\cos^2\alpha+2\cos\alpha\sec\alpha+\sec^2\alpha)\\=\sin^2\alpha+2+\csc^2\alpha+\cos^2\alpha+2+\sec^2\alpha\\=(\sin^2\alpha+\cos^2\alpha)+2+(1+\cot^2\alpha)\\+2+(1+\tan^2\alpha)\,\,[**]\\=1+2+1+\cot^2\alpha+2+1+\tan^2\alpha\\=\tan^2\alpha+\cot^2\alpha+7\quad\text{(proved)}$

Note[**] : $\,\,\csc^2\alpha=1+\cot^2\alpha,\,\,\sec^2\alpha=1+\tan^2\alpha.$

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

$\,\,(xi)\,\,(1+\sec A+\tan A)(1-\csc A+\cot A)=2$

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

Sol. $\quad (1+\sec A+\tan A)(1-\csc A+\cot A)\\=\left(1+\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\times \left(1-\frac{1}{\sin A}+\frac{\cos A}{\sin A}\right)\\=\frac{\cos A+1+\sin A}{\cos A} \times \frac{\sin A-1+\cos A}{\sin A}\\=\frac{1}{\sin A\cos A} \times [(\sin A+\cos A+1)\\ \times (\sin A+\cos A-1)]\\=\frac{1}{\sin A\cos A} \times  [(\sin A+\cos A)^2-1^2]\\=\frac{1}{\sin A\cos A} \times [\sin^2A+\cos^2A\\+2\sin A\cos A-1]\\=\frac{1}{\sin A\cos A} \times [1+2\sin A\cos A-1]\\=\frac{1}{\sin A\cos A} \times 2\sin A\cos A\\=2\quad\text{(proved)}$