Differentiation (Part-38) | S N De

 

Prove the following identities :

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

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

$\,1(ii)\,\,\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1+\sin\theta}{\cos\theta}$

Sol. $\,\,\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\\=\frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\cos\theta}+\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\cos\theta}-\frac{1}{\cos\theta}}\,\,[**]\\=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}\\=\frac{\tan\theta+\sec\theta-(\sec^2\theta-\tan^2\theta)}{1-\sec\theta+\tan\theta}\\=\frac{(\tan\theta+\sec\theta)-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1-\sec\theta+\tan\theta)}\\=\frac{(\tan\theta+\sec\theta)(1-(\sec\theta-\tan\theta))}{(1-\sec\theta+\tan\theta)}\\=\frac{(\tan\theta+\sec\theta)(1-\sec\theta+\tan\theta)}{(1-\sec\theta+\tan\theta)}\\=\tan\theta+\sec\theta\\=\sec\theta+\tan\theta\\=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\\=\frac{1+\sin\theta}{\cos\theta}\,\,\text{(proved)}$

Note[**] : Dividing numerator and denominator by $\,\,\cos\theta.$

$\,1(iii)\,\,\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}$

Sol.  We first calculate, 

$\,\,\frac{1}{\sec A+\tan A}+\frac{1}{\sec A-\tan A}\\=\frac{(\sec A-\tan A)+(\sec A+\tan A)}{(\sec A+\tan A)(\sec A-\tan A)}\\=\frac{2\sec A}{\sec^2A-\tan^2A}\\=\frac{2\sec A}{1}\\=\frac{2}{\cos A}\\=\frac{1}{\cos A}+\frac{1}{\cos A}\\ \text{Hence,}\,\,\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A+\tan A}\\\text{(proved)}$

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

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

$\,\,1(v)\,\,\cos^8A-\sin^8A=(1-2\sin^2A\cos^2A)\\ \times(1-2\sin^2A)$

Sol. $\,\,\cos^8A-\sin^8A\\=(\cos^4A)^2-(\sin^4A)^2\\=(\cos^4A+\sin^4A)(\cos^4A-\sin^4A)\\=[(\cos^2A)^2+(\sin^2A)^2] \\ \times [(\cos^2A)^2-(\sin^2A)^2]\\=[(\cos^2A+\sin^2A)^2-2\sin^2A\cos^2A] \\ \times(\cos^2A+\sin^2A)(\cos^2A-\sin^2A)\\=(1-2\sin^2A\cos^2A)(1-\sin^2A-\sin^2A)\,\,[*]\\=(1-2\sin^2A\cos^2A)(1-2\sin^2A)\,\,\text{(proved)}$

Note[*] : We have used the formula $\,\,\cos^2A+\sin^2A=1.$

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

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

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

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

Hence,  L.H.S.=R.H.S. (proved)

$\,1(vii)\,\cot^2\alpha\frac{\sec\alpha-1}{1+\sin\alpha}+\sec^2\alpha\frac{\sin\alpha-1}{\sec\alpha+1}=0$

Sol. $\,\cot^2\alpha\frac{\sec\alpha-1}{1+\sin\alpha}+\sec^2\alpha\frac{\sin\alpha-1}{\sec\alpha+1}\\=\frac{\cos^2\alpha}{\sin^2\alpha}\times \frac{\frac{1}{\cos\alpha}-1}{1+\sin\alpha}+\frac{1}{\cos^2\alpha}\times \frac{\sin\alpha-1}{\frac{1}{\cos\alpha}+1}\\=\frac{\cos^2\alpha}{\sin^2\alpha}\times \frac{1-\cos\alpha}{\cos\alpha(1+\sin\alpha)}+\frac{1}{\cos^2\alpha} \times \frac{\cos\alpha(\sin\alpha-1)}{1+\cos\alpha}\\=\frac{\cos\alpha}{\sin^2\alpha}\times \frac{(1-\cos\alpha)(1+\cos\alpha)}{(1+\sin\alpha)(1+\cos\alpha)}+\frac{(\sin\alpha-1)}{\cos\alpha(1+\cos\alpha)}\\=\frac{\cos\alpha}{\sin^2\alpha}\times \frac{1-\cos^2\alpha}{(1+\sin\alpha)(1+\cos\alpha)}-\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}\\=\frac{\cos\alpha}{\sin^2\alpha}\times \frac{\sin^2\alpha(1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)(1+\cos\alpha)}-\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}\\=\frac{\cos\alpha(1-\sin\alpha)}{(1-\sin^2\alpha)(1+\cos\alpha)}-\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}\\=\frac{\cos\alpha(1-\sin\alpha)}{\cos^2\alpha(1+\cos\alpha)}-\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}\\=\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}-\frac{1-\sin\alpha}{\cos\alpha(1+\cos\alpha)}\\=0\,\,\,\text{(proved)}$

$\,1(viii)\,\,\,(\cos\alpha\cos\beta+\sin\alpha\sin\beta)^2\\+(\sin\alpha\cos\beta-\cos\alpha\sin\beta)^2=1$

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

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