Question
यदि $\sec \theta = 1\frac{1}{4}$, तो $\tan \frac{\theta }{2} = $
$\sec \theta = \frac{{1 + {{\tan }^2}(\theta /2)}}{{1 - {{\tan }^2}(\theta /2)}}$
$\Rightarrow \frac{5}{4} = \frac{{1 + {{\tan }^2}(\theta /2)}}{{1 - {{\tan }^2}(\theta /2)}}$
$\Rightarrow 5 - 5{\tan ^2}(\theta /2) = 4 + 4{\tan ^2}(\theta /2)$
$\Rightarrow 9{\tan ^2}(\theta /2) = 1\, $
$\Rightarrow \tan (\theta /2) = \frac{1}{3}$.
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