Question

If 0< x < π and cos x + sin x = \(\rm 1\over3\), then cot x is
1. \(\rm {-9 - \sqrt{7} \over 8}\)
2. \(\rm {-9 + \sqrt{7} \over 8}\)
3. \(\rm \frac{8}{ -9 + \sqrt7}\)
4. \(\rm \frac{8}{ -9 - \sqrt7}\)

Answer 1

Correct Answer - Option 4 : \(\rm \frac{8}{ -9 - \sqrt7}\)

Concept:

sin² x + cos² x = 1

\(\rm \sin 2x = \frac{2\tan x}{1 + tan^2 x}\)

tan x = 1/cot x

ax² + bx + c = 0 roots of this quation

\(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Calculation:

Given: cos x + sin x = \(\rm 1\over3\)

Squaring both sides, we get

(cos x + sin x)² = \(\rm ({1\over3})^2\)

⇒ cos² x + sin² x + 2sin x.cos x = \(\rm 1\over 9\)                                (∵ sin2 x + cos2 x = 1 and 2sin x.cos x = sin 2x)

⇒ 1 + sin 2x = \(\rm 1\over 9\)

⇒ sin 2x = \(\rm -{8\over 9}\)

sin x is negative π to 2π 

π < 2x < 2π 

Divide by 2 in the above equation, we get

\(\rm \frac{\pi}{2} < x <\pi\)

⇒ \(\rm -{8\over9} = \frac{2tanx}{1 + tan^2}\)                                                              (∵ \(\rm \sin 2x = \frac{2\tan x}{1 + tan^2 x}\))

⇒ 8tan² x +18 tan x + 8 = 0

⇒ \(\rm tanx = {-18 \pm \sqrt{18^2-4\times 8 \times 8} \over 2\times 8}\)

⇒ \(\rm tanx = {-18 \pm \sqrt{324-256} \over 16}\)

⇒ \(\rm tanx = {-18 \pm \sqrt{68} \over 16}\)

⇒ \(\rm tanx = {-18 \pm 2\sqrt{7} \over 16}\)

⇒ \(\rm tanx = {-9 \pm \sqrt{7} \over 8}\)

⇒ cot x = \(\rm\frac{1}{ tanx} = \frac{8}{ -9 \pm \sqrt7}\)

In, \(\rm \frac{\pi}{2} < x <\pi\) , cot x and tan x are negative

The value of cot x is \(\rm \frac{8}{ -9 - \sqrt7}\)