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Question
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 - x^2/c^2 = 1`
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Solution
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`
`=> zx/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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RELATED QUESTIONS
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= 5(1)
= `square`
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Find the value of sin2θ + cos2θ
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In Δ ABC, ∠ABC = 90°, ∠C = θ°
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But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
