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प्रश्न
If x = a tan θ and y = b sec θ then
विकल्प
`y^2/"b"^2 - x^2/"a"^2` = 1
`x^2/"a"^2 - y^2/"b"^2` = 1
`x^2/"a"^2 + y^2/"b"^2` = 1
`x^2/"a"^2 - y^2/"b"^2` = 0
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उत्तर
`y^2/"b"^2 - x^2/"a"^2` = 1
Explanation;
Hint:
x = a tan θ
`x/"a"` = tan θ
`x^2/"a"^2` = tan2θ
`y^2/"b"^2 - x^2/"a"^2` = sec2θ – tan2θ = 1
y = b sec θ
`y/"b"` = sec θ
`y^2/"b"^2` = sec2θ
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
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(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
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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
