Question

$\sqrt{ \frac{1 + \: sin \: theta}{1 - sin \: theta} } \: is \: equa l \: to$
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Answer 1

$\color{red}{\huge{\underline{\underline{GIVEN\dag}}}}$

• $\sqrt{ \frac{1 + \sin(θ) }{1 - \sin(θ) } }$

$\color{red}{\huge{\underline{\underline{SOLUTION \dag}}}}$

$⇒\sqrt{ \frac{1 + \sin(θ) }{1 - \sin(θ) } } \times \sqrt{ \frac{1 + \sin(θ) }{1 + \sin(θ) } }$

$⇒\sqrt{ \frac{(1 + \sinθ {)}^{2} }{(1) ^{2} - (\sinθ )^{2} } }$

$⇒ \sqrt{ \frac{ {(1 + \sinθ})^{2}}{ \cos {}^{2} (θ) } }$

$⇒ \sqrt{( { \frac{1 + \sin(θ) }{ \cos(θ) } })^{2} }$

$⇒ \frac{1 + \sin(θ) }{ \cos(θ) }$

$⇒ \frac{1}{ \cos(θ) } + \frac{ \sin(θ) }{ \cos(θ) }$

$\color{orange}{\boxed{∴\sqrt{ \frac{1 + \sin(θ) }{1 - \sin(θ) } } = \sec(θ) + \tan(θ)}}$

HOPE THAT HELPS!!