Question

If $cosx-sinx=-\frac{5}{4}$ , where $\frac{\pi }{2} < x < \frac{3 \pi }{4}$ , then $cot\frac{x}{2}$ is equal to

• A. $\frac{4 - \sqrt{7}}{9}$
• B. $8$
• C. $-8$
• D. $\frac{4 + \sqrt{7}}{9}$

Answer 1

Answer: (D)

Given, $4\left(cos x - sin ⁡ x\right)=-5$
$4\left(\frac{1 - t a n^{2} \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}} - \frac{2 tan \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}}\right)=-5$
$\Rightarrow 4-4tan^{2} \frac{x}{2}-8tan ⁡ \frac{x}{2}=-5-5tan^{2} ⁡ \frac{x}{2}$
$\Rightarrow tan^{2}\frac{x}{2}-8tan \frac{x}{2}+9=0$
$\Rightarrow 9cot^{2} \frac{x}{2}-8cot ⁡ \frac{x}{2}+1=0$
$\Rightarrow cot\frac{x}{2}=\frac{4 \pm \sqrt{7}}{9}$
Since, $\frac{\pi }{4} < \frac{x}{2} < \frac{3 \pi }{8}\Rightarrow \sqrt{2}-1 < cot\frac{x}{2} < 1$
$\Rightarrow cot\frac{x}{2}=\frac{4 + \sqrt{7}}{9}$