The sum of roots of the equation tan-11-1-2x-tan-11-1-4x-tan-12-x2 is

The correct option is B 7/3 tan^ 1(2+6x(1+6x+8x^2) 1)=tan^ 1(2x^2) ⇒2+6x6x+8x^2=2x^2 ⇒ 2x^2+6x^3=12x+16x^2 ⇒ 6x^3 14x^2 12x=0 ⇒ x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relation between Roots and Coefficients for Quadratic

The sum of ro...

Question

The sum of roots of the equation $t a n^{- 1} \frac{1}{1 + 2 x} + t a n^{- 1} \frac{1}{1 + 4 x} = t a n^{- 1} \frac{2}{x^{2}}$ is

3 / 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7 / 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9 / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

{tan}^{- 1} \frac{1}{1 + 2 x} + {tan}^{- 1} \frac{1}{4 x + 1} = {tan}^{- 1} \frac{2}{x^{2}}

t a n^{- 1} \frac{1}{2 x + 1} + t a n^{- 1} \frac{1}{4 x + 1} = t a n^{- 1} \frac{2}{x^{2}}

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1}

\sin

x

=

\tan

\sin

x

x \neq \frac{π}{2}

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

{tan}^{- 1} (\frac{1}{1 + 1 + 1^{2}}) + {tan}^{- 1} (\frac{1}{1 + 2 + 2^{2}}) + {tan}^{- 1} (\frac{1}{1 + 3 + 3^{2}}) + . . . . . . . \infty

Join BYJU'S Learning Program