If \( \cos \mathrm{A}=\frac{a}{b} \), prove that \( \frac{\sin A \tan A}{\sec A-1}=\frac{a+1}{b} \)

To prove that \(\frac{\sin A \tan A}{\sec A - 1} = \frac{a + 1}{b}\), we start with the given information that \(\cos A = \frac{a}{b}\). We can find \(\sin A\) using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \implies \sin^2 A = 1 - \cos^2 A \] \[ \sin^2 A = 1 - \left(\frac{a}{b}\right)^2 = 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2} \] Thus, \[ \sin A = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b} \] Next, we compute \(\tan A\) using the definition \(\tan A = \frac{\sin A}{\cos A}\): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{\sqrt{b^2 - a^2}}{b}}{\frac{a}{b}} = \frac{\sqrt{b^2 - a^2}}{a} \] Now, substituting \(\sin A\) and \(\tan A\) in the left-hand side of the expression: \[ \sin A \tan A = \frac{\sqrt{b^2 - a^2}}{b} \cdot \frac{\sqrt{b^2 - a^2}}{a} = \frac{b^2 - a^2}{ab} \] Next, we compute \(\sec A\): \[ \sec A = \frac{1}{\cos A} = \frac{b}{a} \] Thus, \[ \sec A - 1 = \frac{b}{a} - 1 = \frac{b - a}{a} \] Now we can substitute these values back into our original expression: \[ \frac{\sin A \tan A}{\sec A - 1} = \frac{\frac{b^2 - a^2}{ab}}{\frac{b - a}{a}} = \frac{(b^2 - a^2)}{ab} \cdot \frac{a}{(b - a)} = \frac{(b^2 - a^2)}{b(b - a)} \] Now we can factor \(b^2 - a^2\): \[ b^2 - a^2 = (b - a)(b + a) \] This implies: \[ \frac{(b - a)(b + a)}{b(b - a)} = \frac{b + a}{b} \] Thus, we have: \[ \frac{\sin A \tan A}{\sec A - 1} = \frac{b + a}{b} \] Finally, this can be rewritten as: \[ \frac{a + 1}{b} \] Hence, we conclude that: \[ \frac{\sin A \tan A}{\sec A - 1} = \frac{a + 1}{b} \] This proves the required equation. Therefore, we have proven that: \[ \frac{\sin A \tan A}{\sec A - 1} = \frac{a + 1}{b}. \]