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प्रश्न
If x = a tan θ and y = b sec θ then
पर्याय
`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
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Choose the correct alternative:
cos θ. sec θ = ?
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
