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Question
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
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Solution
we get :
`x^2 = (acosθ)^2 = a^2cos^2θ`
`y^2 = (bcotθ)^2 = b^2cot^2θ`
LHS = `a^2/x^2 - b^2/y^2 = a^2/(a^2cos^2θ) - b^2/(b^2 cot^2θ) = 1/(cos^2θ) - 1/cot^2θ`
⇒ LHS = `sec^2θ - tan^2θ = 1 ["Since" 1 + tan^2θ = sec^2θ]`
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