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)
V Vakul

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

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