3. Prove the following identities: a. \( \sec \left(\sec ^{-1} \frac{9}{7}+\sec ^{-1} \frac{11}{7}\right)=99 \)

To prove the identity \( \sec \left(\sec ^{-1} \frac{9}{7}+\sec ^{-1} \frac{11}{7}\right)=99 \), let's define \( x = \sec^{-1} \frac{9}{7} \) and \( y = \sec^{-1} \frac{11}{7} \). This means \( \sec x = \frac{9}{7} \) and \( \sec y = \frac{11}{7} \). Using the addition formula for secant, we have: \[ \sec(x + y) = \frac{\sec x \sec y}{1 - \tan x \tan y} \] We know \( \sec x = \frac{9}{7} \) and \( \sec y = \frac{11}{7} \). We can find \( \tan x \) and \( \tan y \) using the identity \( \tan^2 \theta = \sec^2 \theta - 1 \): \[ \tan^2 x = \left(\frac{9}{7}\right)^2 - 1 = \frac{81}{49} - 1 = \frac{32}{49} \Rightarrow \tan x = \frac{4\sqrt{2}}{7} \] \[ \tan^2 y = \left(\frac{11}{7}\right)^2 - 1 = \frac{121}{49} - 1 = \frac{72}{49} \Rightarrow \tan y = \frac{6\sqrt{2}}{7} \] Now, we substitute these back into the secant addition formula: \[ \tan x \tan y = \frac{4\sqrt{2}}{7} \cdot \frac{6\sqrt{2}}{7} = \frac{48}{49} \] Thus, we have: \[ \sec(x + y) = \frac{\frac{9}{7} \cdot \frac{11}{7}}{1 - \frac{48}{49}} = \frac{\frac{99}{49}}{\frac{1}{49}} = 99 \] Therefore, we conclude that: \[ \sec \left(\sec^{-1} \frac{9}{7} + \sec^{-1} \frac{11}{7}\right) = 99 \] This proves the identity is true!