Question

The number of solutions to the equation$ta{n}^{-1}(x+1)+ta{n}^{-1}x+ta{n}^{-1}(x-1)=\pi -ta{n}^{-1}3$ is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

We have $ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)=\pi -ta{n}^{-1}3-ta{n}^{-1}x=\pi -(ta{n}^{-1}3+ta{n}^{-1}x)$
Taking tan of both sides, we have
$\frac{(x+1)+(x-1)}{1-(x+1)(x-1)}=-\frac{3+x}{1-3x}\Rightarrow 2x-6x^2=(3+x)(x^2-2)\Rightarrow x^3=9x^2-4x-6=0\Rightarrow (x-1)(x^2+10x+6)=0$
$\Rightarrow x=1$ or $x=-5\pm \sqrt{19}$.
Clearly x = 1 satisfies the equation as
L.H.S= $ta{n}^{-1}2+ta{n}^{-1}1+ta{n}^{-1}0=\pi +ta{n}^{-1}\frac{2+1}{1-2}=\pi -ta{n}^{-1}3$ =R.H.S.
While $X=-5\pm \sqrt{19}$ do not satisfy the equation.