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If $\tan \theta + \sec \theta = m$, then prove that $\sec \theta = \frac{m^2+1}{2m}$
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$$\begin{aligned}& \tan \theta + \sec \theta = m \dots (i) \\ & Therefore, \sec \theta - \tan \theta = \frac{1}{m} \dots (ii) \\ & Adding (i)\end{aligned}$$ and $(ii)$ to get
$$\begin{aligned}& 2 \sec \theta = m + \frac{1}{m} \\ & \sec \theta = \frac{m^2+1}{2m}\end{aligned}$$
$$\begin{aligned}& 2 \sec \theta = m + \frac{1}{m} \\ & \sec \theta = \frac{m^2+1}{2m}\end{aligned}$$