Question

Prove that:
2+ sin theta/(cot theta - cosec theta ) = sin theta /(cot theta + cosec theta)​

Answer 1

Proving:

Step-by-step explanation:

$\ Given \ value:\\\\2+\frac{\sin \theta}{\cot\theta-\ cosec \theta}=\frac{\sin \theta}{\cot\theta+\ cosec \theta}$

$\ Solution:\\\\\ L.H.S$

$2+\frac{\sin \theta}{\cot\theta-\ cosec \theta}\\\\\rightarrow 2+\frac{\sin \theta}{\cot\theta-\ cosec \theta} \times \frac{\cot\theta+\ cosec \theta}{\cot\theta+\ cosec \theta}\\\\\rightarrow 2+\frac{\sin \theta(\cot\theta+\ cosec \theta)}{\cot^2\theta-\ cosec^2 \theta} \\\\\rightarrow 2-\frac{\sin \theta(\cot\theta+\ cosec \theta)}{\ cosec^2 \theta-\cot^2\theta} \\\\\therefore \ cosec^2 \theta-\cot^2\theta\\\\\rightarrow 2-\sin \theta(\cot\theta+\ cosec \theta)$

$\rightarrow 2-\sin \theta(\frac{\cos\theta}{\sin \theta}+\frac{1}{\sin \theta})\\\\\rightarrow 2-\sin \theta(\frac{1+\cos \theta}{\sin \theta})\\\\\rightarrow 2-({1+\cos \theta})\\\\\rightarrow 2-1-\cos \theta\\\\\rightarrow 1-\cos \theta$

$\ R.H.S\\\\\frac{\sin \theta}{\cot\theta+\ cosec \theta}\\\\\rightarrow\frac{\sin \theta}{\cot\theta+\ cosec \theta} \times \frac{\cot\theta-\ cosec \theta}{\cot\theta-\ cosec \theta}\\\\\rightarrow \frac{\sin \theta(\cot\theta-\ cosec \theta)}{\cot^2\theta-\ cosec^2 \theta} \\\\\rightarrow -\frac{\sin \theta(\cot\theta-\ cosec \theta)}{\ cosec^2 \theta-\cot^2\theta} \\\\\therefore \ cosec^2 \theta-\cot^2\theta\\\\\rightarrow -\sin \theta(\cot\theta-\ cosec \theta)$

$\rightarrow -\sin \theta(\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta} )\\\\\rightarrow -\sin \theta(\frac{\cos \theta -1 }{\sin \theta})\\\\\rightarrow -(\cos \theta -1)\\\\\rightarrow (1-\cos \theta)$

L.H.S = R.H.S

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