Question

prove this trigonometric eq.​

Answer 1

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To Prove

$\large { \sf{\sqrt{ \frac{1 + sin \theta}{1 - sin \theta} } = sec \theta \: + tan \theta}}$

LHS

$\sf{ \sqrt{ \frac{1 + sin \theta}{1 - sin \theta} } } \\ \\ = \sf{\sqrt{ \frac{1 + sin \theta}{1 \: - sin \theta \: } \times \frac{1 + sin \theta}{1 + sin \theta} } } \\ \\ = \sf{\sqrt{ \frac{(1 + sin \theta) {}^{2} }{1 - sin {}^{2} \theta } } }$

Since,

sin²x + cos²x = 1 » 1 - sin²x = cos²x

$= \sf{ \sqrt{( \frac{1 + sin \theta}{cos \theta} }) {}^{2} } \\ \\ = \sf{\frac{1 + sin \theta}{cos \theta}} \\ \\ = \sf{ \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta} } \\ \\ = \: \huge{ \sf{sec \theta + tan \theta}}$

Hence,proved

Answer 2

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