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if sec theta+tan theta=m,then tan theta is equal to.

$Given sec θ + \tan θ = m \Rightarrow \frac{1}{\cos θ} + \frac{\sin θ}{\cos θ} = m \Rightarrow \frac{1 + \sin θ}{\cos θ} = m - - - - (i) \sec θ - \tan θ = \frac{1}{m} (∵ \sec^{2} θ - \tan^{2} θ = 1) \Rightarrow \frac{1}{\cos θ} - \frac{\sin θ}{\cos θ} = \frac{1}{m} \Rightarrow \frac{1 - \sin θ}{\cos θ} = \frac{1}{m} - - - - (i i) Adding (i) and (ii) we get \Rightarrow \frac{1 + \sin θ}{\cos θ} + \frac{1 - \sin θ}{\cos θ} = m + \frac{1}{m} \Rightarrow \frac{2}{\cos θ} = \frac{m^{2} + 1}{m} \Rightarrow \cos θ = \frac{2 m}{m^{2} + 1} \Rightarrow \sin θ = \frac{m^{2} - 1}{m^{2} + 1} \Rightarrow \tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{m^{2} - 1}{m^{2} + 1}}{\frac{2 m}{m^{2} + 1}} = \frac{m^{2} - 1}{2 m}$

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