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प्रश्न
Differentiate `cot^-1((cos x)/(1 + sinx))` w.r. to x
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उत्तर
Let y = `cot^-1((cos x)/(1 + sinx))`
= `tan^-1((1 + sinx)/(cos x))`
= `tan^-1[(cos^2(x/2) + sin^2(x/2) + 2sin(x/2)cos(x/2))/(cos^2(x/2) - sin^2(x/2))]`
= `tan^-1[{cos(x/2) + sin(x/2)}^2/([cos(x/2) + sin(x/2)][cos(x/2) - sin(x/2)])]`
= `tan^-1[(cos(x/2) + sin(x/2))/(cos(x/2) - sin(x/2))]`
= `tan^-1[(1 + tan(x/2))/(1 - tan(x/2))]`
= `tan^-1[(tan(pi/4) + tan(pi/2))/(1 - tan(pi/4)tan(x/2))]`
= `tan^-1[tan(pi/4 + x/2)]`
∴ y = `pi/4 + x/2`
Differentiating w. r. t. x, we get
`("d"y)/("d"x) = "d"/("d"x)(pi/4 + x/2) = 0 + 1/2 = 1/2`
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