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  4. If π < θ < (3 π /2) and cos θ =-(3/5) , then tan((θ /4)) is equal to

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Q. If $\pi < \theta < \frac{3 \pi }{2}$ and $cos \theta =-\frac{3}{5}$ , then $tan\left(\frac{\theta }{4}\right)$ is equal to

NTA AbhyasNTA Abhyas 2022

Solution:

$cos \theta =-\frac{3}{5}$ $\Rightarrow 2cos^{2}\frac{\theta }{2}-1=-\frac{3}{5}$
$2cos^{2}\frac{\theta }{2}=1-\frac{3}{5}=\frac{2}{5}$
$cos\frac{\theta }{2}=\pm\frac{1}{\sqrt{5}}$
$\Rightarrow cos\frac{\theta }{2}=-\frac{1}{\sqrt{5}}$ $\left(\because \frac{\pi }{2} < \frac{\theta }{2} < \frac{3 \pi }{4}\right)$
Now, $tan^{2}\frac{\theta }{4}=\frac{1 - c o s \frac{\theta }{2}}{1 + c o s \frac{\theta }{2}}=\frac{1 + \frac{1}{\sqrt{5}}}{1 - \frac{1}{\sqrt{5}}}=\frac{\sqrt{5} + 1}{\sqrt{5} - 1}$
$tan^{2}\frac{\theta }{4}=\frac{\left(\sqrt{5} + 1\right)^{2}}{4}=\left(\frac{\left(\sqrt{5} + 1\right)}{2}\right)^{2}$
$\Rightarrow tan\frac{\theta }{4}=\frac{\sqrt{5} + 1}{2}$ $\left(\because \frac{\pi }{4} < \frac{\theta }{4} < \frac{3 \pi }{8}\right)$