Question

$\cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1$

Answer 1

Hint:
We can take the LHS of the given equation. Then by using the trigonometric identities for changing the quadrants and signs of cos in each quadrant, we can simplify each of the terms in the LHS. Then we can expand the bracket and simplify using trigonometric identities to get the RHS.

Complete step by step solution:
We need to prove the trigonometric equation $\cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right] = 1$
We can take the LHS,
$ \Rightarrow LHS = \cos \left( {\dfrac{{3\pi }}{2} + x} \right)\cos \left( {2\pi + x} \right)\left[ {\cot \left( {\dfrac{{3\pi }}{2} - x} \right) + \cot \left( {2\pi + x} \right)} \right]$
We know that trigonometric functions have a period of $2\pi $. Therefore, the value of a trigonometric function of an angle will be the same for the angle $2\pi $added to that angle. So, we can write
$ \Rightarrow \cos \left( {2\pi + x} \right) = \cos \left( x \right)$ … (1)
And similarly, we can write
$ \Rightarrow \cot \left( {2\pi + x} \right) = \cot \left( x \right)$ … (2)
Now we can take \[\cos \left( {\dfrac{{3\pi }}{2} + x} \right)\]
We know that when an odd multiple of $\dfrac{\pi }{2}$ is added to the angle in a trigonometric function, then the function will become the complementary function and the sign will be the sign of the ratio at that quadrant.
In this case, \[\left( {\dfrac{{3\pi }}{2} + x} \right)\] is in the fourth quadrant and we know that cosine ratio is positive in the fourth quadrant.
\[ \Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + x} \right) = \sin x\]…. (3)
Similarly, $\left( {\dfrac{{3\pi }}{2} - x} \right)$ is in the ${3^{rd}}$ quadrant and we know that cot function is positive in the ${3^{rd}}$ quadrant.
\[ \Rightarrow \cot \left( {\dfrac{{3\pi }}{2} - x} \right) = \tan x\] ….. (4)
Now we can substitute equations (1), (2), (3) and (4) in the LHS. So, we get
$ \Rightarrow LHS = \sin x\cos x\left[ {\tan x + \cot x} \right]$
Now we can expand the bracket and multiply the terms.
$ \Rightarrow LHS = \sin x\cos x\tan x + \sin x\cos x\cot x$
We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$ and $\cot x = \dfrac{{\cos x}}{{\sin x}}$.
\[ \Rightarrow LHS = \sin x\cos x\dfrac{{\sin x}}{{\cos x}} + \sin x\cos x \times \dfrac{{\cos x}}{{\sin x}}\]
On cancelling the common terms, we get
\[ \Rightarrow LHS = {\sin ^2}x + {\cos ^2}x\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\]
\[ \Rightarrow LHS = 1\]
We have the RHS also equal to 1.
Therefore, L.H.S = R.H.S
As the LHS and RHS are equal, the equation is proved.

Note:
We can simplify each of the terms of the LHS and simplify it separately to avoid mistakes. While changing the ratio, we must take care of the sign. We arbitrarily assumed that x is an acute angle so that we can find the sign of the values of the functions. While expanding the brackets, we must multiply both the terms outside the brackets to each of the terms inside the bracket.