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In a $ \Delta ABC $, $ \angle B = 90^\circ $, then $ \tan^2\left(\frac{A}{2}\right) $ is

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In a $Δ ABC$, $\angle B = 90circ$, then $\tan(A/2)$ is

Updated On: Apr 15, 2026

$\frac{b-c}{b+c}$
$\frac{b+c}{b-c}$
$\frac{b-2c}{b+c}$
None of these

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The Correct Option is A

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In a \( \Delta ABC \), \( \angle B = 90^\circ \), then \( \tan^2\left(\frac{A}{2}\right) \) is

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The Correct Option is A

Solution and Explanation

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