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If sin^(-1)x-cos^(-1)x=(pi)/(6),"then"...

If `sin^(-1)x-cos^(-1)x=(pi)/(6),"then"`

A

`f(x)=sinx`

B

`f(x)=cosx`

C

`A=(pi)/(4)`

D

`A=(pi)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, D
`int sin^(-1)x cos^(-1)x dx=int [(pi)/(2) sin^(-1)x-(sin^(-1)x)^(2)]dx`
`=(pi)/(2) (x sin^(-1)x+sqrt(1-x^(2)))-(x(sin^(-1)x)^(2)+2sin^(-1)xsqrt(1-x^(2))-2x)+C " (Integrating by parts)" `
`=sin^(-1)x[(pi)/(2)x-x sin^(-1)x-2 sqrt(1-x^(2))]+(pi)/(2)sqrt(1-x^(2))+2x+C`
` :. f^(-1)(x)=sin^(-1)x,f(x)=sinx`
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