Question

If $\tan \left( {\cot x} \right) = \cot \left( {\tan x} \right)$ , then
A.$\sin 2x = \dfrac{4}{{\left( {2n + 1} \right)\pi }}$
B.$\sin 2x = \dfrac{4}{\pi }$
C.$\sin 2x = \dfrac{1}{{\left( {2n + 1} \right)}}$
D.$\sin 2x = n\pi $

Answer 1

Hint: We can change the cot function on the RHS of the equation to tan by using the complementary angle. Then we can take inverses on both sides. Then we can simplify the expression obtained using trigonometric identities. Then we can write the equation in terms of $\sin 2x$ to get the required answer.

Complete step-by-step answer:
We have the equation $\tan \left( {\cot x} \right) = \cot \left( {\tan x} \right)$
We know that $\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot \theta $ .
Here \[\theta = \tan x\] ,
So, the equation will become
$ \Rightarrow \tan \left( {\cot x} \right) = \tan \left( {\dfrac{\pi }{2} - \tan x} \right)$
We know that tan function is periodic in the interval $n\pi $ . So, we can write
$ \Rightarrow \tan \left( {\cot x} \right) = \tan \left( {n\pi + \dfrac{\pi }{2} - \tan x} \right)$
Now we can take tan inverse on both sides,
$ \Rightarrow {\tan ^{ - 1}}\tan \left( {\cot x} \right) = {\tan ^{ - 1}}\tan \left( {n\pi + \dfrac{\pi }{2} - \tan x} \right)$
We know that tan and tan inverse are inverse operations, so we can cancel them.
$ \Rightarrow \cot x = n\pi + \dfrac{\pi }{2} - \tan x$
On taking the terms containing x to one side, we get
$ \Rightarrow \cot x + \tan x = n\pi + \dfrac{\pi }{2}$
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ . So, the LHS will become
$ \Rightarrow \dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin x}}{{\cos x}} = n\pi + \dfrac{\pi }{2}$
On taking the LCM of the LHS, we get
$ \Rightarrow \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\sin x\cos x}} = n\pi + \dfrac{\pi }{2}$
We know that $co{s^2}x + {\sin ^2}x = 1$ . So, the equations will become
$ \Rightarrow \dfrac{1}{{\sin x\cos x}} = n\pi + \dfrac{\pi }{2}$
Now we can divide throughout with 2.
$ \Rightarrow \dfrac{1}{{2\sin x\cos x}} = \dfrac{{n\pi }}{2} + \dfrac{\pi }{4}$
We know that $2\sin x\cos x = \sin 2x$ . So, the denominator will become
$ \Rightarrow \dfrac{1}{{\sin 2x}} = \dfrac{{n\pi }}{2} + \dfrac{\pi }{4}$
We can take the LCM in the RHS.
$ \Rightarrow \dfrac{1}{{\sin 2x}} = \dfrac{{2n\pi + \pi }}{4}$
Taking \[\pi \] common, we get
$ \Rightarrow \dfrac{1}{{\sin 2x}} = \dfrac{{\left( {2n + 1} \right)\pi }}{4}$
On taking reciprocal on both sides, we get
$ \Rightarrow \sin 2x = \dfrac{4}{{\left( {2n + 1} \right)\pi }}$
Therefore, the required relation is $\sin 2x = \dfrac{4}{{\left( {2n + 1} \right)\pi }}$ .
So, the correct answer is option A.

Note: We must change the cot function to tan function using complementary angles, not by taking the reciprocal. We must consider the general solution of tan inverse. As the tan function repeats its value in the period $\pi $ , we can add the term $n\pi $ to the function before taking the inverse. We must convert the tan and cot in terms of sin and cos function for simplification. Before taking the inverse, we must simplify the RHS to one term to avoid errors.