Question

If $3ta{n}^{-1}x+co{t}^{-1}x=\pi ,$ then x equals to

(a) 0 (b) 1 (c) $-1$ (d) $\frac{1}{2}$

Answer 1

(b) Given that, $3ta{n}^{-1}x+co{t}^{-1}x=\pi ,$

$\Rightarrow 2ta{n}^{-1}x+ta{n}^{-1}x+co{t}^{-1}x=\pi \cdots \left(i\right)\Rightarrow 2ta{n}^{-1}x=\pi -\frac{\pi }{2}[\because ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}]\Rightarrow 2ta{n}^{-1}x=\frac{\pi }{2}\Rightarrow ta{n}^{-1}\frac{2x}{1-x^2}=\frac{\pi }{2}[\because 2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2},\forall x\in (-1,1\left)\right]\Rightarrow \frac{2x}{1-x^2}=tan\frac{\pi }{2}\Rightarrow \frac{2x}{1-x^2}=\frac{1}{0}\Rightarrow 1-x^2=0\Rightarrow x^2=1\Rightarrow x=\pm 1\Rightarrow x=1$

Hence, only $x=1$ satisfies the given equation.

Note Here, putting $x=-1$ in the given equation, we get

$3ta{n}^{-1}(-1)+co{t}^{-1}(-1)=\pi \Rightarrow 3ta{n}^{-1}\left[tan\left(\frac{-\pi }{4}\right)\right]+co{t}^{-1}\left[cot\left(\frac{\pi }{4}\right)\right]=\pi \Rightarrow 3ta{n}^{-1}(-tant\frac{\pi }{4})+co{t}^{-1}(-cot\frac{\pi }{4})=\pi \Rightarrow -3ta{n}^{-1}\left(tan\frac{\pi }{4}\right)+\pi -co{t}^{-1}\left(cot\frac{\pi }{4}\right)=\pi \Rightarrow -3.\frac{\pi }{4}+\pi -\frac{\pi }{4}=\pi \Rightarrow -\pi +\pi =\pi \Rightarrow 0\ne \pi$

Hence, $x=-1$ does not satisfy the given equation.