To solve the problem, we need to determine the value of \(\tan A + \sec A\) given that \(\sin A = \frac{1}{\sqrt{10}}\).
- First, we know the identity \(\sin^2 A + \cos^2 A = 1\). Given that \(\sin A = \frac{1}{\sqrt{10}}\), we have:
\(\left(\frac{1}{\sqrt{10}}\right)^2 + \cos^2 A = 1\)
- Thus, \(\frac{1}{10} + \cos^2 A = 1\)
- \(\cos^2 A = 1 - \frac{1}{10} = \frac{9}{10}\)
- Therefore, \(\cos A = \pm \frac{3}{\sqrt{10}}\)
- Next, calculate \(\tan A\), which is:
\(\tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}\)
- Now, calculate \(\sec A\), which is the reciprocal of \(\cos A\):
\(\sec A = \frac{1}{\frac{3}{\sqrt{10}}} = \frac{\sqrt{10}}{3}\)
- Finally, add \(\tan A\) and \(\sec A\):
\(\tan A + \sec A = \frac{1}{3} + \frac{\sqrt{10}}{3} = \frac{1+\sqrt{10}}{3}\)
- Thus, the correct answer is \(\frac{1+\sqrt{10}}{3}\), which corresponds to the first option provided.
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If cos x = p/q and 0° < x < 90°, then the value of tan x is:
