If `sin^(-1)x-cos^(-1)x=pi/6` , then `x=` `1/2` (b) `(sqrt(3))/2` (c) `-1/2` (d) none of these

Explanation

We have given

`sin^(-1)x-cos^(-1)x=pi/6`

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

`pi/2−2⋅cos^(-1)x=pi/6`

`-2⋅cos^(-1)x=pi/6−pi/2`

`-2⋅cos^(-1)x=(pi-3pi)/6`

`-2⋅cos^(-1)x=pi/6-(3pi)/6`

`2⋅cos^(-1)x=(2pi)/6`

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