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प्रश्न
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
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उत्तर
LHS = `sqrt((1 + sin θ)/(1 - sin θ) xx (1 + sin θ)/(1 + sin θ))`
= `sqrt((1 + sin θ)^2/(1 - sin^2θ))`
= `sqrt((1 + sin θ)^2/(cos^2θ)`
= `(1 + sin θ)/cos θ = 1/cos θ + sin θ/cos θ`
= sec θ + tan θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
