Find the exact values

Find the exact values of tan (cos^-1(2/3)) and cos (sin^-1(-3/5)).

tan (cos^-1(2/3))

$$\small{\text{$ \boxed{\tan{ \left(~ \arccos{(x)} ~\right) }=\pm\dfrac {\sqrt{1-x^2}} {x} =\pm\dfrac {\sqrt{1-\left(\dfrac{2}{3}\right)^2}} { \dfrac{2}{3} } =\pm \dfrac{3}{2} \cdot \sqrt{ 1-\dfrac{4}{9} } =\pm \dfrac{3}{2} \cdot \dfrac{ \sqrt{5}}{3} =\pm \dfrac{ \sqrt{5} }{2} } $}}$$

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

$$\small{\text{$ \boxed{\cos{ \left(~ \arcsin{(x)} ~\right) }=\pm\sqrt{1-x^2}=\pm \sqrt{1+ \left( \dfrac{-3}{5} \right)^2}=\pm \dfrac{\sqrt{34} }{ 5 } }$}}$$

Think of a right-angled triangle.  If the cosine of an angle is 2/3, the side adjacent to the angle is proportional to 2 and the hypotenuse is proportional to 3, so the opposite side is proportional to √(3² - 2²) = √5.

Since tan of the angle is opposite/adjacent then tan(cos^-1(2/3)) = (√5)/2

See if you can do something similar with the other one (remembering that in the fourth quadrant sin is negative but cos is positive).

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