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प्रश्न
Differentiate the following w.r.t. x :
`cot^-1[(sqrt(1 + sin ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin ((4x)/3)) - sqrt(1 - sin ((4x)/3)))]`
उत्तर
Let y = `cot^-1[(sqrt(1 + sin ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin ((4x)/3)) - sqrt(1 - sin ((4x)/3)))]`
= `1 + sin ((4x)/3)`
= `1 + cos(pi/2 - (4x)/3)`
= `2cos^2(pi/4 - (2x)/3)`
∴ `sqrt(1 + sin((4x)/3)) = sqrt(2)cos(pi/4 - (2x)/3)`
Also, `1 - sin ((4x)/3)`
= `1 - cos(pi/2 - (4x)/3)`
= `2sin^2(pi/4 - (2x)/3)`
∴ `sqrt(1 - sin((4x)/3)) = sqrt(2)sin(pi/4 - (2x)/3)`
∴ `(sqrt(1 + sin ((4x)/3)) + sqrt(1 - sin ((4x)/3)))/(sqrt(1 + sin((4x)/3) - sqrt(1 - sin((4x)/3)`
= `(sqrt(2)cos(pi/4 - (2x)/3) + sqrt(2)sin(pi/4 - (2x)/3))/(sqrt(2)cos(pi/4 - (2x)/3) - sqrt(2)sin(pi/4 - (2x)/3)`
= `(cos(pi/4 - (2x)/3) + sin(pi/4 - (2x)/3))/(cos(pi/4 - (2x)/3) - sin(pi/4 - (2x)/3)`
= `(1 + tan(pi/4 - (2x)/3))/(1 - tan(pi/4 - (2x)/3)) ...["Dividing by" cos(pi/4 - (2x)/3)`
= `(tan pi/4 + tan(pi/4 - (2x)/3))/(1 - tan pi/4. tan(pi/4 - (2x)/3)) ...[∵ tan pi/4 = 1]`
= `tan[pi/4 + pi/4 - (2x)/3]`
= `tan(pi/2 - (2x)/3)`
= `cot((2x)/3)`
∴ y = `cot^-1[cot((2x)/3)] = (2x)/(3)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"((2x)/3)`
= `(2)/(3)"d"/"dx"(x)`
= `(2)/(3) xx 1`
= `(2)/(3)`
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