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Question
If x = `sqrt(a^(sin^(-1)t))`, y = `sqrt(a^(cos^(-1)t))` show that `dy/dx = - y/x`.
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Solution
Given, x = `sqrt(a^(sin^(-1)t))` and y = `sqrt(a^(cos^(-1)t))`
`dx/dt = 1/2 . 1/(sqrt(a^(sin^(-1)t))). d/dt a ^(sin^(-1)t)`
= `1/2 . 1/ sqrt (a^(sin^(-1)t)). a^(sin^(-1)t) . log a d/dt sin^-1 t`
= `sqrt(a^(sin^(-1)t))/2. log a . 1/ (sqrt(1-t^2)`
`dy/dt = 1/2. 1/ sqrt (a^(cos^(-1)t)). d/dt a^(cos^(-1)t)`
= `1/2 . 1/sqrt (a^(cos^(-1)t)). a^( cos^(-1)t) . log a. (-1)/(sqrt (1 - t^2))`
= `sqrt (a^(cos^(-1)t))/2 .log a (-1)/sqrt(1 - t^2)`
`∴ dy/dx = (dy/dt)/(dx/dt)`
= `(sqrt (a^(cos^(-1)t))/2. log a. (-1)/ sqrt(1 - t^2))/( sqrt (a^(sin^(-1)t))/2 . log a . 1/ sqrt (1 - t^2))`
= `(-sqrt( a^(cos^(-1)t)))/ sqrt (a^(sin^(-1)t))`
= `(-y)/x`
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