Question

solve hurry please...​

Answer 1

Answer:

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

Answer 2

Answer:

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