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प्रश्न
If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`
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उत्तर
`tan^-1[sqrt((1 + cos x)/(1 - cos x))]`
Put `1 + cos x = 2 cos^2 x/2`
`1 - cos x = 2 sin^2 x/2`
∵ `cos x = 2 cos^2 x/2 - 1 = 1 - 2 sin^2 x/2`
= `tan^-1[sqrt((2cos^2 (x/2))/(2sin^2 (x/2)))]`
= `tan^-1 [sqrt(cot^2(x/2))]`
= `tan^-1 [cot(x/2)]`
= `tan^-1 [tan(pi/2- x/2)] ...[cot theta = tan(pi/2 - theta)]`
= `pi/2 - x/2 ...[tan^-1 (tan theta) = theta]`
Differentiating w. r. t. x, we get
`("d"y)/("d"x) = "d"/("d"x) pi/2 - "d"/("d"x) x/2`
∴ `("d"y)/("d"x) = 0 - 1/2 "d"/("d"x) x`
∴ `("d"y)/("d"x) = -1/2`
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