If cosx-sinx=-5/4, where (π)/2<x<(3π)/4, then cotx/2 is equal to

Correct answer: (D) (4+√7)/9. Given, 4cos⁡x-sin⁡x=-5 4(1-tan^2x)/21+tan^2x/2-(2tan⁡x)/21+tan^2x/2=-5 ⇒4-4tan^2⁡x/2-8tan⁡x/2=-5-5tan^2⁡x/2 ⇒tan^2x/2-8tan⁡x/2+9=0 ⇒9cot^2⁡x/2-8cot⁡x/2+1=0 ⇒cotx/2=(4±√7)/9 Since, (π)/4<x/2<(3π)/8⇒√2-1<cotx/2<1 ⇒cotx/2=(4+√7)/9

Q43 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

If $c o s x - s i n x = - \frac{5}{4}$ , where $\frac{π}{2} < x < \frac{3 π}{4}$ , then $c o t \frac{x}{2}$ is equal to

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

Solution

Given,

4 (\cos x - \sin x) = - 5

4 (\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}}) =- 5

\Rightarrow 4 - 4 \tan^{2} \frac{x}{2} - 8 \tan \frac{x}{2} = - 5 - 5 \tan^{2} \frac{x}{2}

\Rightarrow t a n^{2} \frac{x}{2} - 8 \tan \frac{x}{2} + 9 = 0

\Rightarrow 9 \cot^{2} \frac{x}{2} - 8 \cot \frac{x}{2} + 1 = 0

\Rightarrow c o t \frac{x}{2} = \frac{4 \pm \sqrt{7}}{9}

Since,

\frac{π}{4} < \frac{x}{2} < \frac{3 π}{8} \Rightarrow \sqrt{2} - 1 < c o t \frac{x}{2} < 1

\Rightarrow c o t \frac{x}{2} = \frac{4 + \sqrt{7}}{9}

Hence, the correct answer is option: 'D'.

If cosx-sinx=-54, where π2<x<3π4, then cotx2 is equal to

Solution

More Trigonometric Equations Questions

If $c o s x - s i n x = - \frac{5}{4}$ , where $\frac{π}{2} < x < \frac{3 π}{4}$ , then $c o t \frac{x}{2}$ is equal to