Question

find a simplified expression for \\( \tan \left( \cos ^ { - 1 } \left( \frac { t } { 7 } \
ight) \
ight) \\)

Explanation:

Step1: Let $\theta=\cos^{-1}(\frac{t}{7})$

By the definition of inverse cosine, $\cos\theta = \frac{t}{7}$, where $- 7\leq t\leq7$ and $0\leq\theta\leq\pi$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

We get $\sin\theta=\sqrt{1-\cos^{2}\theta}$. Substituting $\cos\theta=\frac{t}{7}$, we have $\sin\theta=\sqrt{1 - (\frac{t}{7})^{2}}=\frac{\sqrt{49 - t^{2}}}{7}$ (since $\sin\theta\geq0$ for $0\leq\theta\leq\pi$).

Step3: Use the formula $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Substituting $\sin\theta=\frac{\sqrt{49 - t^{2}}}{7}$ and $\cos\theta=\frac{t}{7}$, we get $\tan\theta=\frac{\frac{\sqrt{49 - t^{2}}}{7}}{\frac{t}{7}}=\frac{\sqrt{49 - t^{2}}}{t}$ ($t
eq0$). When $t = 0$, $\cos^{-1}(0)=\frac{\pi}{2}$, and $\lim_{t
ightarrow0}\tan(\cos^{-1}(\frac{t}{7}))$ is undefined in the algebraic - substitution sense, but using the right - hand and left - hand limits (considering the domain of $\cos^{-1}x$), we still use the identity - based formula.

Answer:

$\frac{\sqrt{49 - t^{2}}}{t}$