\( f \tan \theta=\frac{3}{7} 1 \) find \( \cos \theta \)

To find \( \cos \theta \) given the equation \( f \tan \theta = \frac{3}{7} \), we first need to express \( \tan \theta \) in terms of \( \sin \theta \) and \( \cos \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] From the equation, we can express \( \tan \theta \) as: \[ \tan \theta = \frac{3}{7f} \] Next, we can use the identity relating \( \sin \theta \), \( \cos \theta \), and \( \tan \theta \): \[ \tan^2 \theta + 1 = \sec^2 \theta \] This can be rewritten in terms of \( \cos \theta \): \[ \frac{\sin^2 \theta}{\cos^2 \theta} + 1 = \frac{1}{\cos^2 \theta} \] Substituting \( \tan \theta \): \[ \left(\frac{3}{7f}\right)^2 + 1 = \frac{1}{\cos^2 \theta} \] Now, let's calculate \( \left(\frac{3}{7f}\right)^2 \): \[ \left(\frac{3}{7f}\right)^2 = \frac{9}{49f^2} \] Substituting this back into the equation gives: \[ \frac{9}{49f^2} + 1 = \frac{1}{\cos^2 \theta} \] Now, we can express \( 1 \) as \( \frac{49f^2}{49f^2} \): \[ \frac{9 + 49f^2}{49f^2} = \frac{1}{\cos^2 \theta} \] Taking the reciprocal: \[ \frac{49f^2}{9 + 49f^2} = \cos^2 \theta \] Finally, taking the square root to find \( \cos \theta \): \[ \cos \theta = \sqrt{\frac{49f^2}{9 + 49f^2}} \] Thus, the expression for \( \cos \theta \) is: \[ \cos \theta = \frac{7f}{\sqrt{9 + 49f^2}} \] This is the final answer for \( \cos \theta \).