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If $0< x< \pi$ and $\cos x+\sin x= 1/2$ then $\tan x$ is

Option 1)
$\frac{\left ( 1-\sqrt{7} \right )}{4}$

Option 2)
$\frac{\left ( 4-\sqrt{7} \right )}{3}$

Option 3)
$-\frac{\left ( 4+\sqrt{7} \right )}{3}$

Option 4)
$\frac{\left ( 1+\sqrt{7} \right )}{4}$

Answers (1)

As we learnt in

Trigonometric Equations -

The equations involving trigonometric function of unknown angles are known as trigonometric equations.

- wherein

e.g. $\cos ^{2}\Theta - 4\cos \Theta = 1$

$\sin x+\cos x= \frac{1}{2}$

$\Rightarrow \sin x = \frac{1}{2}-\cos x$ Square both side we get

$\sin ^2x = \frac{1}{4}+\cos ^2x-\cos x$

$\Rightarrow 1- \cos ^2x=\frac{1}{4}+\cos ^2x-\cos x$ $\Rightarrow 2\cos ^2x-\cos x-\frac{3}{4} =0$

$\Rightarrow \cos x= \frac{1\pm\sqrt{1+6}}{4}$ $=\frac{1\pm\sqrt{7}}{4}$

Now , $\cos x= \frac{1+\sqrt{7}}{4}$ or $\cos x= \frac{1-\sqrt{7}}{4}$

(1) if $\cos x = \frac{1+\sqrt{7}}{4}, \sin x= \frac{1}{2}-\frac{1}{4}-\frac{\sqrt{7}}{4}=\frac{1-\sqrt {7}}{4}<0$

But $x\epsilon \left ( 0,\pi \right ) \Rightarrow \sin x> 0$ Not Possible.

(2) if $\cos x = \frac{1-\sqrt{-7}}{4}, \sin x =\frac{1}{2} -\frac{1}{4}+\frac{\sqrt{7}}{4}$ $= \frac{1+\sqrt{7}}{4}$

Here

$\tan x= \frac{\sqrt{7+1}}{1-\sqrt{7}}$ $= - \frac{ \left ( \sqrt{7+1} \right )\sqrt{7+1}}{\left ( 7-1 \right )}$ $= - \:\frac{\left ( 7+1+2\sqrt{7} \right )}{6}$ $= - \frac{\left ( 4+\sqrt{7} \right )}{3}$

Option 1)

$\frac{\left ( 1-\sqrt{7} \right )}{4}$

Incorrect

Option 2)

$\frac{\left ( 4-\sqrt{7} \right )}{3}$

Incorrect

Option 3)

$-\frac{\left ( 4+\sqrt{7} \right )}{3}$

Correct

Option 4)

$\frac{\left ( 1+\sqrt{7} \right )}{4}$

Incorrect

Posted by

Vakul

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