Question

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

Answer 1

In this problem, we need to find the value of $\sin 2x$ where we are provided with a trigonometric expression which is $\tan \left( {\cot x} \right) = \cot \left( {\tan x} \right)$. For this, first we will use the identity $\cot \left( \alpha \right) = \tan \left( {\dfrac{\pi }{2} - \alpha } \right)$. Then, we will use the result $\tan x = \tan \alpha \Rightarrow x = n\pi + \alpha$. Then, we will use some basic trigonometric identities.

Complete step-by-step solution:
In this problem, it is given that $\tan \left( {\cot x} \right) = \cot \left( {\tan x} \right) \cdots \cdots \left( 1 \right)$.
Now we will use the identity $\cot \left( \alpha \right) = \tan \left( {\dfrac{\pi }{2} - \alpha } \right)$ on RHS of equation $\left( 1 \right)$.
Therefore, we get $\tan \left( {\cot x} \right) = \tan \left( {\dfrac{\pi }{2} - \tan x} \right) \cdots \cdots \left( 2 \right)$.
Now we are going to use the result $\tan x = \tan \alpha \Rightarrow x = n\pi + \alpha$ in equation $\left( 2 \right)$.
Therefore, we get $\cot x = n\pi + \dfrac{\pi }{2} - \tan x \cdots \cdots \left( 3 \right)$.
Let us simplify the equation $\left( 3 \right)$. Therefore, we get $\cot x + \tan x = n\pi + \dfrac{\pi }{2} \cdots \cdots \left( 4 \right)$.
Now we need to find the value of $\sin 2x$.
Therefore, we will convert the equation $\left( 4 \right)$ in terms of sine and cosines.
For this, we will use some basic trigonometric identities. That is, we will use identities $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ on LHS of equation $\left( 4 \right)$.

Therefore, we get $\dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin x}}{{\cos x}} = \dfrac{{2n\pi + \pi }}{2} \cdots \cdots \left( 5 \right)$
Let us simplify the LHS and RHS of the equation $\left( 5 \right)$.
Therefore, we get $\dfrac{{\left( {\cos x} \right)\left( {\cos x} \right) + \left( {\sin x} \right)\left( {\sin x} \right)}}{{\left( {\sin x} \right)\left( {\cos x} \right)}} = \dfrac{{\left( {2n + 1} \right)\pi }}{2}$
$\Rightarrow \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\left( {\sin x} \right)\left( {\cos x} \right)}} = \dfrac{{\left( {2n + 1} \right)\pi }}{2} \cdots \cdots \left( 6 \right)$
Now we are going to use the Pythagorean identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ on the LHS of equation $\left( 6 \right)$.
Therefore, we get $\dfrac{1}{{\left( {\sin x} \right)\left( {\cos x} \right)}} = \dfrac{{\left( {2n + 1} \right)\pi }}{2}$
$\Rightarrow \left( {\sin x} \right)\left( {\cos x} \right) = \dfrac{2}{{\left( {2n + 1} \right)\pi }} \cdots \cdots \left( 7 \right)$
Let us multiply by the number $2$ on both sides of equation $\left( 7 \right)$.
Therefore, we get
$\Rightarrow2\left( {\sin x} \right)\left( {\cos x} \right) = \dfrac{4}{{\left( {2n + 1} \right)\pi }} \cdots \cdots \left( 8 \right)$
Now we are going to use the formula $2\sin \theta \cos \theta = \sin 2\theta$ on the LHS of equation $\Rightarrow \left( 8 \right)$. Therefore, we get $\sin 2x = \dfrac{4}{{\left( {2n + 1} \right)\pi }}$

Hence the correct answer is option A .

Note: There are various distinct trigonometric identities. When trigonometric functions are involved in an equation then trigonometric identities are useful to solve that equation. We can use identities $\cos e{c^2}x - {\cot ^2}x = 1$ and ${\sec ^2}x - {\tan ^2}x = 1$ to solve many trigonometric problems. These identities are called Pythagorean identities.