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, $4\left(cosx-sinx\right)=-5$
$4\left(\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}-\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}\right)=-5$
$\Rightarrow 4-4ta{n}^2\frac{x}{2}-8tan\frac{x}{2}=-5-5ta{n}^2\frac{x}{2}$
$\Rightarrow ta{n}^2\frac{x}{2}-8tan\frac{x}{2}+9=0$
$\Rightarrow 9co{t}^2\frac{x}{2}-8cot\frac{x}{2}+1=0$
$\Rightarrow cot\frac{x}{2}=\frac{4\pm \sqrt{7}}{9}$
Since, $\frac{\pi }{4}<\frac{x}{2}<\frac{3\pi }{8}\Rightarrow \sqrt{2}-1<cot\frac{x}{2}<1$
$\Rightarrow cot\frac{x}{2}=\frac{4+\sqrt{7}}{9}$