How do you evaluate $tan ({cos}^{- 1} (- \frac{2}{3}))$ $tan(cos^-1(-2/3))$ without a calculator?

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Answer 1 · 2016-08-21T05:56:08

$- \frac{\sqrt{5}}{2}$ $-sqrt 5/2$

Let $a = cos & (- 1) (- \frac{2}{3}) \in Q 2$ $a = cos&(-1)(-2/3) in Q2$ .

Tangent is negative in Q2, and sine is positive..

Now, $cos a = - \frac{2}{3}$ $cos a =-2/3$ and, the given expression is

$tan a = \frac{sin a}{cos a} . = \frac{\sqrt{1 - {cos}^{2} a}}{cos a}$ $tan a = sin a/cos a.=sqrt(1-cos^2a)/cos a$

$= \frac{\sqrt{1 - {(- \frac{2}{3})}^{2}}}{- \frac{2}{3}}$ $=sqrt(1-(-2/3)^2)/(-2/3)$

$= - \frac{\sqrt{5}}{2}$ $=-sqrt 5/2$

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