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

Answer 1

Hint: 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.
\[\begin{align}
  & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{2(3)}{1-{{(3)}^{2}}} \right) \right)+\dfrac{\sqrt{10}}{\cos \left( {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right) \right)}+1=0 \\
 & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{-3}{4} \right) \right)+\dfrac{\sqrt{10}}{\left( \dfrac{1}{\sqrt{10}} \right)}+1=0 \\
 & \Rightarrow (-12)\left( \dfrac{-3}{4} \right)+10+1=0 \\
 & \Rightarrow 20=0 \\
\end{align}\]
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.
\[\begin{align}
  & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{2(3)}{1-{{(3)}^{2}}} \right) \right)-\dfrac{\sqrt{10}}{\cos \left( {{\cos }^{-1}}\left( \dfrac{1}{\sqrt{10}} \right) \right)}+1=0 \\
 & \Rightarrow -12Tan\left( Ta{{n}^{-1}}\left( \dfrac{-3}{4} \right) \right)-\dfrac{\sqrt{10}}{\left( \dfrac{1}{\sqrt{10}} \right)}+1=0 \\
 & \Rightarrow (-12)\left( \dfrac{-3}{4} \right)-10+1=0 \\
 & \Rightarrow 0=0 \\
\end{align}\]
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.