Question

tan theta + sin theta upon 10 theta minus sin theta is equal to cosec theta + cot theta square​

Answer 1

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Answer 2

$\frac{tan \theta + sin \theta}{tan \theta - sin \theta} \\ = \frac{sin \theta( \frac{1}{cos \theta } + 1)}{sin \theta( \frac{1}{cos \theta} - 1)} \\ = \frac{1 + cos \theta}{1 - cos \theta} \\ rationalizing \: the \: denominator \\ = \frac{1 + cos \theta}{1 - cos \theta} \times \frac{1 + cos \theta}{1 + cos \theta} \\ = \frac{( {1 + cos \theta})^{2} }{ {sin}^{2} \theta} \\ = ( { \frac{1 + cos \theta}{sin \theta} })^{2} \\ = ( {cosec \theta + cot \theta})^{2}$