## Thomas' Calculus 13th Edition

$\cos{x}=-\dfrac{4}{5}$ $\tan{x}= -\dfrac{3}{4}$
Since $x$ lies in the second quadrant, $\cos{x}$ is negative. Using the identity: $$\cos{x}=-\sqrt{1-\sin^2{x}}$$ $$\cos{x}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}$$ $$\cos{x}=-\dfrac{4}{5}$$ $\because \tan{x}=\dfrac{\sin{x}}{\cos{x}}$ $\therefore \tan{x}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}= -\dfrac{3}{4}$