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Differentiate w.r.t. x the function: cot-1[1+sinx+1-sinx1+sinx-1-sinx], 0<x<π2 - Mathematics

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प्रश्न

Differentiate w.r.t. x the function:

`cot^(-1) [(sqrt(1+sinx) + sqrt(1-sinx))/(sqrt(1+sinx) - sqrt(1-sinx))]`, ` 0 < x < pi/2`

योग

उत्तर

Let, y = `cot^-1 [(sqrt(1 + sin x) + sqrt(1 - sin x))/(sqrt(1 + sin x) - sqrt(1 - sin x))]`

Now, `1 + sin x = sin^2  x/2  + cos^2  x/2 + 2 sin  x/2 cos  x/2`

`= (cos  x/2 + sin  x/2)`

`therefore sqrt(1 + sin x) = cos  x/2 + sin  x/2`

Similarly,

`sqrt(1 + sin x) = cos  x/2 + sin  x/2`

y = `cot^-1 [((cos  x/2 + sin  x/2) + (cos  x/2 - sin  x/2))/((cos  x/2 + sin  x/2) - (cos  x/2 + sin  x/2))]`

`= cot^-1  [(cos  x/2 + sin  x/2 + cos  x/2 - sin  x/2)/(cos  x/2 + sin  x/2 - cos  x/2 + sin  x/2)]`

`= cot^-1  [(2 cos  x/2)/(2  sin  x/2)]`

`= cot^-1  (cot  x/2)`

y = `x/2`

On differentiating with respect to x,

`dy/dx = 1/2 * d/dx (x) = 1/2`

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अध्याय 5: Continuity and Differentiability - Exercise 5.9 [पृष्ठ १९१]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.9 | Q 6 | पृष्ठ १९१

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