Question

If $X=n\pi -{{\tan }^{-1}}(3)$ is a solution of the equation $12\tan 2x+\dfrac{\sqrt{10}}{\cos x}+1=0$ if

& (\text{A) n is any integer} \\\ & \text{(B) n is odd integer} \\\ & \text{(C) n is a positive integer} \\\ & \text{(D) n}\in \text{2M,M}\in \text{I} \\\ \end{aligned}$$

Answer 1

Let us assume $X=n\pi -{{\tan }^{-1}}(3)$ as equation (1). Now let’s assume $12\tan 2x+\dfrac{\sqrt{10}}{\cos x}+1=0$ as equation (2). Now we will substitute equation (1) and equation (2). Now we have to assume two cases and we have to proceed further. Let us assume n is even as Case-1. We know that if n is even , then $\tan \left( n\pi -x \right)=-\tan x$ and $\cos (n\pi -x)=\cos x$. So, by using these relations we will find whether the condition is satisfied or not. Let us assume n is odd as Case-2. We know that if n is odd, then $\tan \left( n\pi -x \right)=-\tan x$ and $\cos (n\pi -x)=-\cos x$.So, by using these relations, we will find whether the condition is satisfied or not.

Complete step-by-step answer:
Now let us assume
$X=n\pi -{{\tan }^{-1}}(3)......(1)$
$12\tan 2x+\dfrac{\sqrt{10}}{\cos x}+1=0.......(2)$
Now let us substitute equation (1) in equation (2).
$12\tan 2(n\pi -{{\tan }^{-1}}(3))+\dfrac{\sqrt{10}}{\cos \left( n\pi -{{\tan }^{-1}}(3) \right)}+1=0$
Case-1: Let us assume n is an even integer.
If n is even, then $\tan \left( n\pi -x \right)=-\tan x$ and $\cos (n\pi -x)=\cos x$.
So, by using above relations, we get
$\Rightarrow -12(\tan (2{{\tan }^{-1}}(3)))+\dfrac{\sqrt{10}}{\cos \left( {{\tan }^{-1}}(3) \right)}+1=0$
We know that $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and ${{\tan }^{-1}}\dfrac{a}{b}={{\cos }^{-1}}\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ , here we have a=3, b=1.

& \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{2(3)}{1-{{(3)}^{2}}} \right) \right)+\dfrac{\sqrt{10}}{\cos \left( {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right) \right)}+1=0 \\\ & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{-3}{4} \right) \right)+\dfrac{\sqrt{10}}{\left( \dfrac{1}{\sqrt{10}} \right)}+1=0 \\\ & \Rightarrow (-12)\left( \dfrac{-3}{4} \right)+10+1=0 \\\ & \Rightarrow 20=0 \\\ \end{aligned}$$ But this is never possible. So, we cannot say that n is even. Case 2: Let us assume n is an odd integer. If n is odd, then $$\tan \left( n\pi -x \right)=-\tan x$$ and $$\cos (n\pi -x)=-\cos x$$. So, by using above relations, we get $$\Rightarrow -12(\tan (2{{\tan }^{-1}}(3)))-\dfrac{\sqrt{10}}{\cos \left( {{\tan }^{-1}}(3) \right)}+1=0$$ We know that $$2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$$ and $${{\tan }^{-1}}\dfrac{a}{b}={{\cos }^{-1}}\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$$ , here we have a=3, b=1. $$\begin{aligned} & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{2(3)}{1-{{(3)}^{2}}} \right) \right)-\dfrac{\sqrt{10}}{\cos \left( {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right) \right)}+1=0 \\\ & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{-3}{4} \right) \right)-\dfrac{\sqrt{10}}{\left( \dfrac{1}{\sqrt{10}} \right)}+1=0 \\\ & \Rightarrow (-12)\left( \dfrac{-3}{4} \right)-10+1=0 \\\ & \Rightarrow 0=0 \\\ \end{aligned}$$ So, we can say that n is odd. **So, the correct answer is “Option B”.** **Note:** While solving this problem, we should not have any misconceptions. We should remember that if n is odd, then $$\tan \left( n\pi -x \right)=-\tan x$$ and $$\cos (n\pi -x)=-\cos x$$, If n is even, then $$\tan \left( n\pi -x \right)=-\tan x$$ and $$\cos (n\pi -x)=\cos x$$. But some students will have a misconception that if n is even, then $$\tan \left( n\pi -x \right)=-\tan x$$ and $$\cos (n\pi -x)=-\cos x$$, If n is odd, then $$\tan \left( n\pi -x \right)=-\tan x$$ and $$\cos (n\pi -x)=\cos x$$. This will give us option D as correct. So, we should be careful while applying this concept.