If \tan \theta+\sec \theta=3, then what is the value of 3 \tan \theta+9 \sec \theta?

Explanation

Since \sec^2\theta - \tan^2\theta = 1, we have \sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta} = \frac{1}{3}. Adding the two equations gives 2\sec\theta = \frac{10}{3} \implies \sec\theta = \frac{5}{3}. Subtracting them gives 2\tan\theta = \frac{8}{3} \implies \tan\theta = \frac{4}{3}. The value of 3\tan\theta + 9\sec\theta = 3(\frac{4}{3}) + 9(\frac{5}{3}) = 4 + 15 = 19.

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