find the five remaining trigonometric functions of \\( \\alpha \\).\n\\( \\tan \\alpha=-\\frac{1}{4}…

find the five remaining trigonometric functions of \\( \\alpha \\).\n\\( \\tan \\alpha=-\\frac{1}{4}, \\alpha \\) in quadrant ii\n\\( \\cot \\alpha= \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)\n\\( \\sec \\alpha= \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

find the five remaining trigonometric functions of \\( \\alpha \\).\n\\( \\tan \\alpha=-\\frac{1}{4}, \\alpha \\) in quadrant ii\n\\( \\cot \\alpha= \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)\n\\( \\sec \\alpha= \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Answer

Explanation:

Step1: Use the identity (1+\tan^{2}\alpha=\sec^{2}\alpha)

Given (\tan\alpha =-\frac{1}{4}), then (\tan^{2}\alpha=\left(-\frac{1}{4}\right)^{2}=\frac{1}{16}). Substitute into the identity: (1 + \frac{1}{16}=\sec^{2}\alpha), so (\sec^{2}\alpha=\frac{16 + 1}{16}=\frac{17}{16}).

Step2: Determine the sign of (\sec\alpha)

Since (\alpha) is in quadrant II, (\cos\alpha<0) (because in quadrant II, (x) - coordinate is negative and (\cos\alpha=\frac{x}{r}), (r>0)). And (\sec\alpha=\frac{1}{\cos\alpha}), so (\sec\alpha<0). Take the square - root of (\sec^{2}\alpha=\frac{17}{16}), (\sec\alpha=-\sqrt{\frac{17}{16}}=-\frac{\sqrt{17}}{4})

Answer:

(\sec\alpha =-\frac{\sqrt{17}}{4})