Prove that:
$\dfrac{{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}}{{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}} = \tan 6x$

Hint: For a question like this we approach the solution by simplifying anyone the side and proving it equal to the other side, here also we will simplify the left-hand side using some of the trigonometric formulas like
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
We simplify in such a manner that it results in the equivalent value to the other side expression

Complete step by step Answer:

Given data:$\dfrac{{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}}{{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}} = \tan 6x$
Taking the expression in the left-hand side
$ \Rightarrow \dfrac{{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}}{{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}}$
Using the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$ in the numerator
$ \Rightarrow \dfrac{{2\sin \left( {\dfrac{{7x + 5x}}{2}} \right)\cos \left( {\dfrac{{7x - 5x}}{2}} \right) + 2\sin \left( {\dfrac{{9x + 3x}}{2}} \right)\cos \left( {\dfrac{{9x - 3x}}{2}} \right)}}{{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}}$
And now using the formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$in the denominator
$ \Rightarrow \dfrac{{2\sin \left( {\dfrac{{7x + 5x}}{2}} \right)\cos \left( {\dfrac{{7x - 5x}}{2}} \right) + 2\sin \left( {\dfrac{{9x + 3x}}{2}} \right)\cos \left( {\dfrac{{9x - 3x}}{2}} \right)}}{{2\cos \left( {\dfrac{{7x + 5x}}{2}} \right)\cos \left( {\dfrac{{7x - 5x}}{2}} \right) + 2\cos \left( {\dfrac{{9x + 3x}}{2}} \right)\cos \left( {\dfrac{{9x - 3x}}{2}} \right)}}$
On simplification we get,
$ \Rightarrow \dfrac{{2\sin \left( {6x} \right)\cos \left( x \right) + 2\sin \left( {6x} \right)\cos \left( {3x} \right)}}{{2\cos \left( {6x} \right)\cos \left( x \right) + 2\cos \left( {6x} \right)\cos \left( {3x} \right)}}$
Taking \[2sin\left( {6x} \right)\] common from the numerator, we get,
$ \Rightarrow \dfrac{{2\sin \left( {6x} \right)[\cos \left( x \right) + \cos \left( {3x} \right)]}}{{2\cos \left( {6x} \right)\cos \left( x \right) + 2\cos \left( {6x} \right)\cos \left( {3x} \right)}}$
Now, taking \[2cos\left( {6x} \right)\] common from the denominator, we get,
$ \Rightarrow \dfrac{{2\sin \left( {6x} \right)[\cos \left( x \right) + \cos \left( {3x} \right)]}}{{2\cos \left( {6x} \right)[\cos \left( x \right) + \cos \left( {3x} \right)]}}$
Dividing both numerator and the denominator by $2[\cos \left( x \right) + \cos \left( {3x} \right)]$, we get,
$ \Rightarrow \dfrac{{\sin \left( {6x} \right)}}{{\cos \left( {6x} \right)}}$
Now using the formula $\dfrac{{\sin \left( A \right)}}{{\cos \left( A \right)}} = \tan A$
$ \Rightarrow \tan 6x$, which is equal to the left-hand side in the given equation
Since, Left-hand side=right-hand side
We have proved the given equation

Note: We also prove that $\dfrac{{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}}{{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}} = \tan 6x$, by substituting the value of x
Substituting \[x = 0\]
Left-hand side$ = \dfrac{{(\sin 0 + \sin 0) + (\sin 0 + \sin 0)}}{{(\cos 0 + \cos 0) + (\cos 0 + \cos 0)}}$
$ = 0$, since $\sin 0 = 0$
right-hand side$ = \tan 6(0)$
\[ = 0\] , $\tan 0 = 0$
Since, Left-hand side=right-hand side\[ = 0\]
We have proved the given equation, but do not attempt this type of solution for the descriptive type question, this substitution method is just a way to check or verification.