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If $sec A = \frac{9}{4}$ then what is the value of cot A ? |
$\frac{4}{\sqrt{65}}$ $\frac{9}{\sqrt{65}}$ $\frac{\sqrt{65}}{9}$ $\frac{\sqrt{65}}{4}$ |
$\frac{4}{\sqrt{65}}$ |
We are given that, sec A = \(\frac{9}{4}\) { using , sec A = \(\frac{H}{B}\) } By using pythagoras theorem , P² + B² = H² P² + 4² = 9² P² = 81 - 16 = 65 P = \(\sqrt { 65}\) Now, cotA = \(\frac{B}{P}\) = \(\frac{4}{√65}\) |
