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प्रश्न
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
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उत्तर
\[\boxed{\text{sec}^2θ} = 1 + \text{tan}^2θ\] ...[Fundamental trigonometric identity]
\[\boxed{\text{sec}^2θ} - \text{tan}^2θ = 1\]
(sec θ + tan θ) . (sec θ – tan θ) = \[\boxed{1}\]
`sqrt(3) . (sectheta - tan theta)` = 1
\[(\text{sec} θ - \text{tan} θ) = \boxed{\frac{1}{\sqrt3}}\]
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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
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
