We are given that \( \cos A = \frac{4}{5} \). To find \( \tan A \), we can use the following trigonometric identity: \[ \tan A = \frac{\sin A}{\cos A} \] First, we need to find \( \sin A \). Using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \] Substitute \( \cos A = \frac{4}{5} \) into the equation: \[ \sin^2 A + \left(\frac{4}{5}\right)^2 = 1 \] \[ \sin^2 A + \frac{16}{25} = 1 \] \[ \sin^2 A = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \] \[ \sin A = \frac{3}{5} \] Now, we can calculate \( \tan A \): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]
The correct option is (C): \(\frac{4}{3}\)
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.