Calculus: Early Transcendentals 8th Edition

For example, let $x=1$ LHS = $\tan^{-1}1=\frac{\pi}{4}$ RHS = $\frac{\sin^{-1}1}{\cos^{-1}1}=\frac{\frac{\pi}{2}}{0}=$undefined Since LHS$\neq$RHS, the statement is not true.