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प्रश्न
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
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उत्तर
LHS = `sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1))`
= `(sqrt( secθ - 1) sqrt( secθ - 1) + sqrt( secθ + 1)sqrt( secθ + 1))/(sqrt(secθ - 1)sqrt(secθ + 1))`
= `((sqrt( secθ - 1))^2 + (sqrt( secθ + 1))^2)/(sqrt(secθ - 1)sqrt(secθ + 1))`
= `(secθ - 1 + secθ + 1)/(sqrt(sec^2 - 1))`
= `(2secθ)/sqrt(tan^2θ)`
= `(2secθ)/(tanθ)`
= `(2 1/cosθ)/(sinθ/cosθ)`
= `(2 1/sinθ)`
= 2 cosecθ.
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