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Prove that cot^−1(√(1+sinx)+√(1−sinx)/√(1+sinx)−√(1−sinx))=x/2; x ∈ (0,π/4) - Mathematics

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Prove that `cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=x/2;x in (0,pi/4) `

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Solution

`cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))`

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

 

`=cot^(-1)(sqrt((cos(x/2)+sin(x/2))^2+sqrt((cos(x/2)-sin(x/2))^2))/(sqrt((cos(x/2)+sin(x/2))^2)-sqrt((cos(x/2)-sin(x/2))^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) {((cos(x/2)+sin(x/2))+(cos(x/2)-sin(x/2)))/((cos(x/2)+sin(x/2))-(cos(x/2)-sin(x/2)))}   [∵0<x<pi/4⇒cos(x/2)>sin (x/4)]`

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

`=cot^(-1)(cotx/2)`

`=x/2`

`=RHS`

Hence proved

Concept: Properties of Inverse Trigonometric Functions
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