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प्रश्न
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
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उत्तर
Given:
`x = a sec theta cos phi`
`=> x/a = sec theta cos phi` ........(1)
`y = b sec theta sin phi`
`=> y/b = sec theta sin phi`
`=> y/b = sec theta sin phi`
`=> z/c = tan theta`
We have to prove that `x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
Squaring the above equations and then subtracting the third from the sum of the first two, we have
`(x/a)^2 + (y/b)^2 - (z/c)^2 = (sec theta cos phi)^2 + (sec theta sin phi)^2 - (tan theta)^2`
`=> x^2/ a^2 + y^2/b^2 - z^2/c62 = sec^2 theta cos^2 phi + sec^2 theta sin^2 phi - tan^2 theta`
`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = (sec^2 theta cos^2 phi + sec^2 theta sin&2 phi) - tan^2 theta`
`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = sec^2 theta(cos^2 phi + sin^2 phi) - tan^2 theta`
`=> x^2/a^2 + y^2/b^2 - z^2/c^2= sec^2 theta (1) = tan^2 theta`
`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = sec^2 theta - tan^2 theta`
`=> x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
Hence proved.
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
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= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
