# Chapter 4 - Review Exercises - Page 644: 27

The values of the remaining trigonometric function are, $\cos t=\frac{\sqrt{21}}{7},\text{ }\tan t=\frac{2\sqrt{3}}{3},\text{ }\csc t=\frac{\sqrt{7}}{2}\text{, }\sec t=\frac{\sqrt{21}}{3},\text{ }\cot t=\frac{\sqrt{3}}{2}$

#### Work Step by Step

Consider the provided trigonometric functions: $\sin t=\frac{2}{\sqrt{7}}$ According to Pythagorean Identities: ${{\sin }^{2}}t+{{\cos }^{2}}t=1$ Now, substitute the value of $\sin t$ in the equation ${{\sin }^{2}}t+{{\cos }^{2}}t=1$. Therefore, ${{\left( \frac{2}{\sqrt{7}} \right)}^{2}}+{{\cos }^{2}}t=1$ The square of, ${{\left( \frac{2}{\sqrt{7}} \right)}^{2}}$ is $\left( \frac{4}{7} \right)$, Therefore, $\frac{4}{7}+{{\cos }^{2}}t=1$ Subtract, $\frac{4}{7}$ from the sides, \begin{align} & -\frac{4}{7}+\frac{4}{7}+{{\cos }^{2}}t=1-\frac{4}{7} \\ & {{\cos }^{2}}t=1-\frac{4}{7} \end{align} On further simplification, \begin{align} & {{\cos }^{2}}t=1-\frac{4}{7} \\ & {{\cos }^{2}}t=\frac{3}{7} \\ \end{align} Take the square root on both sides, $\cos t=\sqrt{\frac{3}{7}}$ This implies that $\cos t=\sqrt{\frac{3}{7}}=\frac{\sqrt{21}}{7}$ The value of $\cos t$ is positive in the given interval $0\le t\le \frac{\pi }{4}$. Hence, the value of $\cos t$ is $\frac{\sqrt{21}}{7}$. Now, the value of $\tan t$ can be obtained by using the values of $\sin t$ and $\cos t$. Therefore, $\tan t=\frac{y}{x}$ Substitute the value of $\sin t$ and $\cos t$ in the function $\tan t=\frac{y}{x}$. $\tan t=\frac{\frac{2}{\sqrt{7}}}{\sqrt{\frac{3}{7}}}$ This implies that $\tan t=\frac{2\sqrt{3}}{3}$ Hence, the value of $\tan t$ is $\frac{2\sqrt{3}}{\sqrt{3}}$ Now, for $\csc t$ as: $\csc t=\frac{1}{\sin t}$ Substitute, the values of $\sin t$ in $\csc t=\frac{1}{\sin t}$. $\csc t=\frac{1}{\frac{2}{\sqrt{7}}}$ This implies that $\csc t=\frac{\sqrt{7}}{2}$ Hence, the value of $\csc t$ is $\frac{\sqrt{7}}{2}$ Now, for $\sec t$ as: $\sec t=\frac{1}{\cos t}$ Substitute, the value of $\cos t$ in $\sec t=\frac{1}{\cos t}$, \begin{align} & \sec t=\frac{1}{\frac{\sqrt{21}}{7}} \\ & =\frac{7}{\sqrt{21}} \end{align} This implies that $\sec t=\frac{\sqrt{21}}{3}$ Hence, the value of $\sec t$ is $\frac{\sqrt{21}}{3}$ . Now, for $\cot t$ as: $\cot t=\frac{x}{y}$ Substitute the value of $\sin t$ and $\cos t$ in the function $\cot t=\frac{x}{y}$. $\cot t=\frac{\frac{2}{\sqrt{7}}}{\frac{\sqrt{21}}{7}}$ This implies that $\cot t=\frac{\sqrt{3}}{2}$ Hence, the value of $\cot t$ is $\frac{\sqrt{3}}{2}$ .

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