Question

Please give step by step solution

Answer 1

The final answer is L.H.S ≠ R.H.S

Step-by-step explanation:

$\ Given \ value:\\\\\frac{\sec \theta - 1}{\sec \theta + 1} = (\cot \theta - \cosec \theta )^2 \\\\\ Solution:\\\\\ Solve \ L.H.S \\\\\frac{\sec \theta - 1}{\sec \theta + 1}\\\\\rightarrow \frac{\sec \theta - 1}{\sec \theta + 1} \times \frac{\sec \theta - 1}{\sec \theta - 1}\\\\\rightarrow \frac{(\sec \theta - 1)^2}{(\sec \theta)^2 - (1)^2} \\\\\rightarrow \frac{(\sec^2 \theta +1 - 2 \sec \theta)}{\sec^2 \theta - 1}$

$\rightarrow \frac{(\sec^2 \theta +1 - 2 \sec \theta)}{1+ \tan^2 \theta - 1} \\\\\rightarrow \frac{(\sec^2 \theta +1 - 2 \sec \theta)}{\tan^2 \theta} \\\\\rightarrow \frac{\sec^2 \theta}{\tan^2 \theta} +\frac{1}{\tan^2 \theta} - 2 \frac{\sec \theta}{\tan^2 \theta} \\\\\rightarrow \frac{1}{\cos^2 \theta}\times \frac{\cos^2 \theta}{\sin^2 \theta} +\frac{1}{\tan^2 \theta} - 2 \frac{1}{\cos \theta} \times \frac{\cos^2 \theta}{\sin^2 \theta}$

$\rightarrow \frac{1}{\sin^2 \theta} +\frac{1}{\tan^2 \theta} - 2 \frac{\cos \theta}{\sin^2 \theta} \\\\\rightarrow \ cosec^2 \theta + \cot^2 \theta - 2 \cot \theta\ cosec \theta \\\\\rightarrow (\ cosec \theta- \cot \theta )^2\\\\\ taking \ minus \ (- ) \ as \ a \ common \\\\\rightarrow - (\cot \theta - \ cosec \theta)^2 \\\\\ That's \ why \ L.H.S \ \neq \ R.H.S$

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• Proving: https://brainly.in/question/8112701