Question

the angle t is an acute angle and sin t and cos t are given. use identities to find tan t, csc t, sec t, and cot t. where necessary, rationalize denominators.

\\( \sin t = \frac { 5 } { 7 } , \cos t = \frac { 2 \sqrt { 6 } } { 7 } \\)

\\( \tan t = \frac { 5 \sqrt { 6 } } { 12 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

\\( \csc t = \frac { 7 } { 5 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

\\( \sec t = \frac { 7 \sqrt { 6 } } { 12 } \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

\\( \cot t = \square \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Recall the cotangent identity

The identity for cotangent is \(\cot t=\frac{\cos t}{\sin t}\) (or also \(\cot t = \frac{1}{\tan t}\), but using \(\frac{\cos t}{\sin t}\) is direct here).

Step2: Substitute the given values

We know that \(\sin t=\frac{5}{7}\) and \(\cos t = \frac{2\sqrt{6}}{7}\). Substituting these into the formula for \(\cot t\), we get:

$$ \cot t=\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}} $$

Step3: Simplify the fraction

When dividing fractions, we multiply by the reciprocal of the denominator. So \(\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}\). The 7 in the numerator and denominator cancels out, leaving \(\frac{2\sqrt{6}}{5}\)? Wait, no, wait. Wait, no, let's check again. Wait, no, the identity is \(\cot t=\frac{\cos t}{\sin t}\), but also \(\cot t=\frac{1}{\tan t}\). We know that \(\tan t=\frac{5\sqrt{6}}{12}\), so \(\cot t=\frac{1}{\frac{5\sqrt{6}}{12}}=\frac{12}{5\sqrt{6}}\). Now rationalize the denominator: multiply numerator and denominator by \(\sqrt{6}\), so \(\frac{12\sqrt{6}}{5\times6}=\frac{2\sqrt{6}}{5}\)? Wait, no, wait, let's do it with \(\cos t\) and \(\sin t\) again. \(\cos t=\frac{2\sqrt{6}}{7}\), \(\sin t = \frac{5}{7}\). So \(\frac{\cos t}{\sin t}=\frac{2\sqrt{6}/7}{5/7}=\frac{2\sqrt{6}}{5}\)? Wait, but also, since \(\tan t=\frac{5\sqrt{6}}{12}\), then \(\cot t=\frac{1}{\tan t}=\frac{12}{5\sqrt{6}}\). Let's rationalize \(\frac{12}{5\sqrt{6}}\): multiply numerator and denominator by \(\sqrt{6}\), we get \(\frac{12\sqrt{6}}{5\times6}=\frac{2\sqrt{6}}{5}\). Wait, but let's check the values again. Wait, the given \(\tan t=\frac{5\sqrt{6}}{12}\), so \(\cot t\) should be the reciprocal. So \(\frac{12}{5\sqrt{6}}\), rationalizing: multiply numerator and denominator by \(\sqrt{6}\), numerator becomes \(12\sqrt{6}\), denominator becomes \(5\times6 = 30\), so \(\frac{12\sqrt{6}}{30}=\frac{2\sqrt{6}}{5}\). Alternatively, using \(\cos t\) and \(\sin t\): \(\frac{2\sqrt{6}/7}{5/7}=\frac{2\sqrt{6}}{5}\). Yes, that's correct. Wait, but wait, maybe I made a mistake earlier. Wait, no, let's re - express:

Wait, \(\cot t=\frac{\cos t}{\sin t}=\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{7}\times\frac{7}{5}=\frac{2\sqrt{6}}{5}\). Yes, that's correct. The 7s cancel, and we are left with \(\frac{2\sqrt{6}}{5}\)? Wait, no, wait, no! Wait, the given \(\tan t=\frac{5\sqrt{6}}{12}\), so \(\cot t\) is the reciprocal, so \(\frac{12}{5\sqrt{6}}\). Let's rationalize \(\frac{12}{5\sqrt{6}}\):

Multiply numerator and denominator by \(\sqrt{6}\):

Numerator: \(12\times\sqrt{6}=12\sqrt{6}\)

Denominator: \(5\times\sqrt{6}\times\sqrt{6}=5\times6 = 30\)

Then \(\frac{12\sqrt{6}}{30}=\frac{2\sqrt{6}}{5}\). Yes, that's correct. So \(\cot t=\frac{2\sqrt{6}}{5}\)? Wait, but let's check with the other method. Since \(\cot t=\frac{\cos t}{\sin t}\), substituting \(\cos t=\frac{2\sqrt{6}}{7}\) and \(\sin t=\frac{5}{7}\), we have \(\frac{2\sqrt{6}/7}{5/7}=\frac{2\sqrt{6}}{5}\). So that's the value of \(\cot t\).

Wait, but wait, maybe I messed up the identity. Wait, \(\tan t=\frac{\sin t}{\cos t}\), so \(\cot t=\frac{\cos t}{\sin t}\), which is correct. So with \(\sin t = 5/7\) and \(\cos t=2\sqrt{6}/7\), then \(\cot t=\frac{2\sqrt{6}/7}{5/7}=\frac{2\sqrt{6}}{5}\).

Answer:

\(\frac{2\sqrt{6}}{5}\)