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प्रश्न
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
विकल्प
ab
a2 − b2
a2 + b2
a2 b2
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उत्तर
Given:
`x= a secθ, y=b tanθ`
So,
`b^2x^2-a^2 y^2`
=` b^2(a secθ)^2-a^2(btan θ)^2`
= `b^2 a^2 sec^2 θ-a^2 b^2 tan^2θ`
=` b^2 a^2 (sec^2θ-tan^2 θ)`
We know that,`
`sec^2θ-tan^2θ=1`
Therfore,
`b^2x^2-a^2y^2=a^2b^2`
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Solution:
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= `(625 - 49)/49`
= `square/49`
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