Question

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

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\frac{\sqrt{5}+1}{2}$
• C. $\frac{-\sqrt{5}+1}{4}$
• D. $\frac{-\sqrt{5}-1}{4}$

Answer 1

Answer: (B)

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