Trigonometry & Inverse Trigonometry Question - None

Question

Let $\sec x + \tan x = \frac{22}{7}$, where $0 < x < \frac{\pi}{2}$. The value of $\tan\left(\frac{x}{2}\right)$ is

Accepted Answer

Step 1: Use the identity $\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} = \frac{22}{7}$.

Step 2: Use half-angle substitution: $\tan\left(\frac{x}{2}\right) = t$, then $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$.

Step 3: Substitute to get $\frac{1 + \frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}} = \frac{22}{7}$.

Step 4: Simplify: $\frac{1 + t^2 + 2t}{1 - t^2} = \frac{22}{7}$ → $\frac{(1+t)^2}{(1-t)(1+t)} = \frac{22}{7}$ → $\frac{1+t}{1-t} = \frac{22}{7}$.

Step 5: Solve: $7(1+t) = 22(1-t)$ → $7 + 7t = 22 - 22t$ → $29t = 15$ → $t = \frac{15}{29}$.

∴ Answer is A.

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