How to prove \[\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x\]?

Hint: We need to remember here the basic rules of trigonometry first, mainly the formula of sum to product identities. These include formulas for \[\sin x + \sin y\], \[\sin x - \sin y\], \[\cos x + \cos y\], \[\cos x - \cos y\]. By applying these formulas, we can easily solve this question.

Complete step-by-step solution:
To prove: \[\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x\]
Proof:
First, we need to take here the LHS and RHS:
LHS\[ = \dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}}\]
RHS\[ = \cot 2x\]
First, we will try to solve LHS. In LHS also, we will try to solve the numerator first. The numerator is:
\[\Rightarrow \sin x - \sin 3x + \sin 5x - \sin 7x\]
First, we will try to make groups. We will group the first two and the last two terms, and we get:
\[ \Rightarrow (\sin x - \sin 3x) + (\sin 5x - \sin 7x)\]
Now, we will use the trigonometric formula:
\[\sin a - \sin b = 2\cos \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}\]
Now, we will put the values according to the formula, and we get:
\[ \Rightarrow (\sin x - \sin 3x) + (\sin 5x - \sin 7x) = \left( {2\cos \dfrac{{x + 3x}}{2}\sin \dfrac{{x - 3x}}{2}} \right) + \left( {2\cos \dfrac{{5x + 7x}}{2}\sin \dfrac{{5x - 7x}}{2}} \right)\]
Now, we will just simplify, and we get:
\[ \Rightarrow \left( {2\cos \dfrac{{4x}}{2}\sin \dfrac{{ - 2x}}{2}} \right) + \left( {2\cos \dfrac{{12x}}{2}\sin \dfrac{{ - 2x}}{2}} \right)\]
Now, we will cancel the terms that are similar and which are divisible, and we get:
\[ \Rightarrow \left( {2\cos (2x)\sin ( - x)} \right) + \left( {2\cos (6x)\sin ( - x)} \right)\]
When we open the brackets, we get:
\[\Rightarrow 2\cos (2x)\sin ( - x) + 2\cos (6x)\sin ( - x)\]
Now, we will try to remove the negative sign, and we get:
\[ \Rightarrow - 2\cos (2x)\sin (x) - 2\cos (6x)\sin (x)\]
Now, we will take out the common term \[ - 2\sin x\], and we get:
\[\Rightarrow - 2\sin x(\cos (2x) + \cos (6x))\]
Similarly, we will solve the denominator as well. The denominator is:
\[ \Rightarrow \cos x - \cos 3x - \cos 5x + \cos 7x\]
First, we will try to make groups. We will group the first two and the last two terms, and we get:
\[\Rightarrow (\cos x - \cos 3x) - (\cos 5x - \cos 7x)\]
Now, we will use the trigonometric formula:
\[\Rightarrow \cos x - \cos y = - 2\sin \dfrac{{x + y}}{2}\sin \dfrac{{x - y}}{2}\]
Now, we will put the values according to the formula, and we get:
\[ \Rightarrow (\cos x - \cos 3x) - (\cos 5x - \cos 7x) = \left( { - 2\sin \dfrac{{x + 3x}}{2}\sin \dfrac{{x - 3x}}{2}} \right) + \left( { - 2\sin \dfrac{{5x + 7x}}{2}\sin \dfrac{{5x - 7x}}{2}} \right)\]
Now, we will just simplify, and we get:
\[\Rightarrow \left( { - 2\sin \dfrac{{4x}}{2}\sin \dfrac{{ - 2x}}{2}} \right) + \left( { - 2\sin \dfrac{{12x}}{2}\sin \dfrac{{ - 2x}}{2}} \right)\]
Now, we will cancel the terms that are similar and which are divisible, and we get:
\[ \Rightarrow \left( { - 2\sin (2x)\sin ( - x)} \right) + \left( { - 2\sin (6x)\sin ( - x)} \right)\]
When we open the brackets, we get:
\[\Rightarrow - 2\sin (2x)\sin ( - x) - 2\sin (6x)\sin ( - x)\]
Now, we will try to remove the negative sign, and we get:
\[ \Rightarrow 2\sin (2x)\sin (x) + 2\sin (6x)\sin (x)\]
Now, we will take out the common term \[2\sin x\], and we get:
\[\Rightarrow 2\sin x(\sin (2x) + \sin (6x))\]
Now, when we rewrite our LHS, we get:
LHS\[ \Rightarrow \dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \dfrac{{ - 2\sin x(\cos (2x) + \cos (6x))}}{{2\sin x(\sin (2x) + \sin (6x))}}\]
Now, we can cancel the similar terms, and we get:
\[ \Rightarrow \dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \dfrac{{ - (\cos (2x) + \cos (6x))}}{{(\sin (2x) + \sin (6x))}}\]
Now, we will use the trigonometric formulas:
\[\cos a + \cos b = 2\cos \dfrac{{x + y}}{2}\cos \dfrac{{x - y}}{2}\]
\[\sin a - \sin b = 2\cos \dfrac{{x + y}}{2}\sin \dfrac{{x - y}}{2}\]
Now, we will put the values according to the formula, and we get:
\[\Rightarrow \dfrac{{ - (\cos (2x) + \cos (6x))}}{{(\sin (2x) + \sin (6x))}} = \dfrac{{ - \left( {2\cos \dfrac{{2x + 6x}}{2}\cos \dfrac{{2x - 6x}}{2}} \right)}}{{\left( {2\cos \dfrac{{2x + 6x}}{2}\sin \dfrac{{2x - 6x}}{2}} \right)}}\]
\[\Rightarrow \dfrac{{ - \left( {2\cos (4x)\cos ( - 2x)} \right)}}{{\left( {2\cos (4x)\sin ( - 2x)} \right)}}\]
Now, the similar terms get cancelled out, and we get:
\[\Rightarrow \dfrac{{ - \left( {2\cos ( - 2x)} \right)}}{{\left( {2\sin ( - 2x)} \right)}}\]
\[\Rightarrow \dfrac{{ - \left( {\cos ( - 2x)} \right)}}{{\left( {\sin ( - 2x)} \right)}}\]
We know that \[\cos ( - \theta ) = \cos \theta \]. So, we get:
\[ \Rightarrow \dfrac{{ - \cos 2x}}{{ - \sin 2x}}\]
\[ \Rightarrow \cot 2x\]
LHS\[ = \cot 2x\]

Hence, LHS=RHS (hence proved). Therefore \[\dfrac{{\sin x - \sin 3x + \sin 5x - \sin 7x}}{{\cos x - \cos 3x - \cos 5x + \cos 7x}} = \cot 2x\] is proved.

Note: We need to keep a check on both the LHS and RHS while solving. In this question, when we were solving our LHS, we had to check that in what terms our RHS is present, and then according to that we need to change our LHS to prove the question. Here in the question we have used we have few trigonometric formulas to solve the problem like \[\sin a - \sin b = 2\cos \dfrac{{a + b}}{2}\sin \dfrac{{a - b}}{2}\] , \[\Rightarrow \cos x - \cos y = - 2\sin \dfrac{{x + y}}{2}\sin \dfrac{{x - y}}{2}\].