If \tan x.\tan y = p and x+y = \frac{\pi}{6}, then \tan x and \tan y satisfy the equation

  • Option 1)

    x^{2}-\sqrt{3}(1-p)x+a=0

  • Option 2)

    \sqrt3x^{2}-(1-p)x+p\sqrt3=0

  • Option 3)

    x^{2}+\sqrt{3}(1+p)x-p=0

  • Option 4)

    \sqrt{3}x^{2}+(1+p)x-p\sqrt{3}=0

 

Answers (1)
A Aadil

 

Subtraction Formulae -

\tan \left ( A-B \right )= \frac{\tan A-\tan B}{1+\tan A\tan B}

- wherein

A and B are two angles.

 

 \tan x\cdot \tan y = p

x+y = \frac{\pi }{6}

\tan (x+y) = \frac{\tan x+\tan y}{1-\tan x+\tan y}

\tan \left ( \frac{\pi }{6} \right )= \tan 30^{\circ} = \frac{\tan x+\tan y}{1-p}

\therefore \frac{1-p}{\sqrt{3}} = \tan x+\tan y

\therefore x^{2} = x (\tan x+\tan y) + \tan x+\tan y=0

\therefore x^{2}-x\cdot \left ( \frac{1-p}{\sqrt{3}} \right )+p=0

\therefore \sqrt{3}x^{2}- x(1-p)+p\sqrt{3}=0

 


Option 1)

x^{2}-\sqrt{3}(1-p)x+a=0

This option is incorrect.

Option 2)

\sqrt3x^{2}-(1-p)x+p\sqrt3=0

This option is correct.

Option 3)

x^{2}+\sqrt{3}(1+p)x-p=0

This option is incorrect.

Option 4)

\sqrt{3}x^{2}+(1+p)x-p\sqrt{3}=0

This option is incorrect.

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