Question

tan[sin^-1(-1 / 2)]

Answer 1

So if \(x=\sin^{-1}\left(-\dfrac{1}{2}\right)\), what is \(\tan x\) ?

Answer 2

\[\tan \sin^{-1} \left( \frac{ -1 }{ 2 } \right)=\tan \left( \frac{- \pi }{6 } \right)=-\tan \frac{ \pi }{ 6 }\]
\[=-\frac{ \frac{ \pi }{ 6 } }{\cos \frac{ \pi }{ 6 } }=-\frac{ \frac{ 1 }{ 2 } }{\frac{ \sqrt{3} }{ 2 } }\]