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प्रश्न
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
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उत्तर
`"LHS" = sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ))`
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get,
`"LHS" = sqrt((1 + sin θ)/(1 - sin θ) × (1 + sin θ)/(1 + sin θ)) + sqrt((1 - sin θ)/(1 + sin θ) × (1 - sin θ)/(1 - sin θ))`
`"LHS" = sqrt((1 + sin θ)^2/(1 - sin^2 θ)) + sqrt((1 - sin θ)^2/(1 - sin^2 θ))`
`"LHS" = sqrt((1 + sin^2θ)/(1 - sin^2 θ)) + sqrt((1 - sin^2θ)/(1 - sin^2 θ))`
`"LHS" = sqrt((1 + sin^2θ)/(cos^2 θ)) + sqrt((1 - sin^2θ)/(cos^2 θ))`
`"LHS" = (1 + sin θ)/(cos θ) + (1 - sin θ)/(cos θ)`
`"LHS" = (1 + cancel(sin θ) + 1 -cancel(sin θ))/(cos θ)`
LHS = `2/(cos θ)`
LHS = 2. `1/(cos θ)`
LHS = 2. sec θ
RHS = 2. sec θ
LHS = RHS
Hence proved.
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