Question

root 1 + sin theta by 1 minus sin theta + root 1 minus sin theta + by 1 + sin theta is equal to 2 Sin Theta

Answer 1

$\sqrt{ \frac{1 + \sin(x) }{1 - \sin(x) } } = \sqrt{ (\frac{1 + \sin(x) }{1 - \sin(x) }) ( \frac{1 + \sin(x) }{1 - \sin(x) }) } \\ = \sqrt{ \frac{(1 + \sin(x))^{2} }{ \cos(x) ^{2} } } = \frac{1 + \sin(x) }{ \cos(x) } \\ similarly \: \sqrt{ \frac{1 - \sin(x) }{1 + \sin(x) } } = \frac{1 - \sin(x) }{ \cos(x) } \\ hence \: \frac{1 + \sin(x) + 1 - \sin(x) }{ \cos(x) } = 2 \sec(x)$
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