`cos(sin^-1(-3/5))`

Explanation

Here, we will use,

`sin^-1(-x) = -sin^-1(x) and cos(-x) = cosx`

Now,

`cos(sin^-1(-3/5)) = cos(-sin^-1(3/5)) `

`=cos(sin^-1(3/5))`

`=cos(cos^-1(sqrt(1-(3/5)^2)))`

`=cos(cos^-1(sqrt(16/25)))`

`=cos(cos^-1 (4/5))`

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