Question

How do you solve and find the value of `tan(sin^-1(1/2))`?

Answer 1

`tan(sin^(-1)(1/2))=1/sqrt3`

Explanation:

Inverse trigonometric ratios are the inverse functions of the trigonometric functions. As an example, `y=sinx` means sine of `/_x` is `y`. As an inverse trigonometric ration this means that `arcsiny=x` or `sin^(-1)y=x`.

As `sin(pi/6)=1/2`, we have `sin^(-1)(1/2)=pi/6`

and hence

`tan(sin^(-1)(1/2))=tan(pi/6)=1/sqrt3`