Find the exact value of the expression, if possible. Do not use a calculator.
-$$\tan \left( \sec ^ { - 1 } \frac { \sqrt { x ^ { 2 } + 9 } } { x } \right)$$
A) $\frac { 3 } { x }$
B) $\frac { x \sqrt { x ^ { 2 } + 9 } } { x ^ { 2 } + 9 }$
C) $3 x$
D) $\frac { \sqrt { x ^ { 2 } + 3 } } { x ^ { 2 } + 3 }$

Question

Quizplus · Accepted Answer

To find the exact value of $\tan \left( \sec ^ { - 1 } \frac { \sqrt { x ^ { 2 } + 9 } } { x } \right)$, we first interpret the expression inside the tangent function. The expression $\sec ^ { - 1 } \frac { \sqrt { x ^ { 2 } + 9 } } { x }$ represents the angle whose secant is $\frac { \sqrt { x ^ { 2 } + 9 } } { x }$. Secant is the reciprocal of cosine, so we are looking for an angle in a right triangle where the hypotenuse over the adjacent side is $\frac { \sqrt { x ^ { 2 } + 9 } } { x }$. This implies a triangle where the adjacent side (next to the angle) is $x$, the hypotenuse (opposite the right angle) is $\sqrt { x ^ { 2 } + 9 }$, and using the Pythagorean theorem, the opposite side (opposite our angle) must be $3$, because $x^2 + 3^2 = x^2 + 9$.The tangent of an angle is the ratio of the opposite side to the adjacent side in a right triangle, so for this triangle, $\tan$ of our angle is $\frac{3}{x}$.

Find the Exact Value of the Expression, If Possible tan⁡(sec⁡−1x2+9x)\tan \left( \sec ^ { - 1 } \frac { \sqrt { x ^ { 2 } + 9 } } { x } \right)tan(sec−1xx2+9​​)

Find the Exact Value of the Expression, If Possible $\tan \left( \sec ^ { - 1 } \frac { \sqrt { x ^ { 2 } + 9 } } { x } \right)$