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प्रश्न
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
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उत्तर
LHS = `1/(sec θ - tan θ)`
= `1/((1/cos θ) - (sin θ/cos θ))`
= `(cos θ xx (1 + sin θ))/((1 - sin θ) xx ( 1 + sin θ))`
= `(cos θ( 1 + sin θ))/(1 - sin^2 θ)`
= `(cos θ( 1 + sin θ))/(cos^2 θ)`
= `1/cos θ + sin θ/cos θ`
= sec θ + tan θ
= RHS
Hence proved.
संबंधित प्रश्न
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`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
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
