Question

Prove LHS=RHS class 10 maths question Trignometry

Answer 1

Answer:

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