Question

answer the question ​

Answer 1

Answer:

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Step-by-step explanation:

ars5d6f8houe5e5dygiy8ufhftsearcugudtwrs

Answer 2

Answer:

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