Question

Please help I will mark as brainliest​

Answer 1

Answer:

$\sqrt{ \frac{1 - sinθ}{1 +sinθ } } = secθ - tanθ \\ \sqrt{ \frac{1 - sinθ}{1 + sinθ} } \times \sqrt{ \frac{1 - sinθ}{1 -sinθ } } = secθ- tanθ \\ \sqrt{ \frac{(1 - sinθ)(1 - sinθ)}{(1 +sinθ )(1 - sinθ)} } = secθ - tanθ \\ \sqrt{ \frac{ {(1 - sinθ)}^{2} }{ {(1)}^{2} - {(sinθ)}^{2} } } =sec θ - tanθ \\ \sqrt{ \frac{ {(1 - sin)}^{2} }{1 - {sin}^{2}θ } } = secθ - tanθ \\ \sqrt{ \frac{ {( 1- sinθ)}^{2} }{ {cos}^{2}θ } } =secθ - tanθ \\ \frac{1 - sinθ}{cosθ} = secθ- tanθ \\ \frac{1}{cosθ} - \frac{sinθ}{cosθ}= secθ - tanθ \\ secθ-tan θ= secθ - tanθ \\ LHS=RHS \\ Hence \: proved$