First slide

Multiple and sub- multiple Angles

Question

If $0 < A < \frac{π}{6}$ and $\sin A + \cos A = \frac{\sqrt{7}}{2}$ , then $\tan \frac{A}{2} =$

Moderate

A

$\frac{\sqrt{7} - 2}{3}$
B

$\frac{\sqrt{7} + 2}{3}$
C

$\frac{\sqrt{7}}{3}$
D

none of these

Solution

We have,
$\begin{array}{l} \sin A + \cos A = \frac{\sqrt{7}}{2} \\ \Rightarrow \frac{2 \tan \frac{A}{2}}{1 + \tan^{2} \frac{A}{2}} + \frac{1 - \tan^{2} \frac{A}{2}}{1 + \tan^{2} \frac{A}{2}} = \frac{\sqrt{7}}{2} \\ \Rightarrow 4 \tan \frac{A}{2} + 2 - 2 \tan^{2} \frac{A}{2} = \sqrt{7} + \sqrt{7} \tan^{2} \frac{A}{2} \\ \Rightarrow (\sqrt{7} + 2) \tan^{2} \frac{A}{2} - 4 \tan \frac{A}{2} + (\sqrt{7} - 2) = 0 \\ \Rightarrow \tan \frac{A}{2} = \frac{4 \pm \sqrt{16 - 4 (\sqrt{7} + 2) (\sqrt{7} - 2)}}{2 (\sqrt{7} + 2)} \end{array}$
$\begin{array}{l} \Rightarrow \tan \frac{A}{2} = \frac{4 \pm 2}{2 (\sqrt{7} + 2)} \\ \Rightarrow \tan \frac{A}{2} = \frac{3}{\sqrt{7} + 2}, \frac{1}{\sqrt{7} + 2} \\ \Rightarrow \tan \frac{A}{2} = \sqrt{7} - 2, \frac{\sqrt{7} - 2}{3} \\ \Rightarrow \tan \frac{A}{2} = \frac{\sqrt{7} - 2}{3} [∵ 0 < A < π / 6 ∴ \tan \frac{A}{2} < 1] \end{array}$

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