Write the following as an algebraic expression in u, u 0.
-$\tan \left( \cos ^ { - 1 } \frac { \mathrm { u } } { 3 } \right)$
A) $\frac { u \sqrt { 9 - u ^ { 2 } } } { 9 - u ^ { 2 } }$
B) $\frac { \sqrt { u ^ { 2 } - 9 } } { u }$
C) $\frac { \sqrt { 9 - u ^ { 2 } } } { u }$
D) $\mathrm { u } ^ { 2 } \sqrt { 9 }$

Question

Write the following as an algebraic expression in u, u > 0.
-$\tan \left( \cos ^ { - 1 } \frac { \mathrm { u } } { 3 } \right)$
A) $\frac { u \sqrt { 9 - u ^ { 2 } } } { 9 - u ^ { 2 } }$
B) $\frac { \sqrt { u ^ { 2 } - 9 } } { u }$
C) $\frac { \sqrt { 9 - u ^ { 2 } } } { u }$
D) $\mathrm { u } ^ { 2 } \sqrt { 9 }$

Quizplus · Accepted Answer

To find the algebraic expression for $\tan \left( \cos ^ { - 1 } \frac { u } { 3 } \right)$, we use the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Here, $\cos(\theta) = \frac{u}{3}$, so we need to find $\sin(\theta)$. Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we get $\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{u}{3}\right)^2} = \sqrt{\frac{9 - u^2}{9}}$. Thus, $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\sqrt{9 - u^2}/3}{u/3} = \frac{\sqrt{9 - u^2}}{u}$.

Write the Following as an Algebraic Expression in U, U tan⁡(cos⁡−1u3)\tan \left( \cos ^ { - 1 } \frac { \mathrm { u } } { 3 } \right)tan(cos−13u​)

Write the Following as an Algebraic Expression in U, U $\tan \left( \cos ^ { - 1 } \frac { \mathrm { u } } { 3 } \right)$