Prove that \[\dfrac{{{\text{sin5x + sinx - 2sin3x}}}}{{{\text{cos5x - cosx}}}} = \tan x\].

Hint: In this problem we utilize the formulae we have already learnt in chapter compound angles of Trigonometry. The formulae are
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos 2x = 1 - 2{\sin ^2}x$
Here we combine the first two terms of the numerator to apply the first mentioned above. Then we extract common terms in the numerator. In the numerator we will have cos2x which is simplified by applying the third formula mentioned above. We now simplify the denominator of the question by applying the second formula mentioned above. We will cancel the terms common in both numerator and denominator. Now, the problem is in simple form and the rest can be solved. Let us try it!

Complete step-by-step solution:
In this question we have to prove that the value of \[\dfrac{{{\text{sin5x + sinx - 2sin3x}}}}{{{\text{cos5x - cosx}}}} = \tan x\]
$\dfrac{{{\text{sin5x + sinx - 2sin3x}}}}{{{\text{cos5x - cosx}}}} $
$= \dfrac{{\left( {{\text{sin5x + sinx}}} \right){\text{ - 2sin3x}}}}{{{\text{cos5x - cosx}}}} $
Here we use the formula of summation of two trigonometric functions as $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) $
$= \dfrac{{2\sin \left( {\dfrac{{5x + x}}{2}} \right)\cos \left( {\dfrac{{5x - x}}{2}} \right) - 2\sin 3x}}{{\cos 5x - \cos x}} $
$= \dfrac{{2\sin \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right) - 2\sin 3x}}{{\cos 5x - \cos x}} $
$= \dfrac{{2\sin 3x\cos 2x - 2\sin 3x}}{{\cos 5x - \cos x}} $
Now we take the common part $2\sin 3x$ from the numerator expression.
$= \dfrac{{2\sin 3x\left( {\cos 2x - 1} \right)}}{{\cos 5x - \cos x}} $
Now we use the trigonometric identity $\cos 2x = 1 - 2{\sin ^2}x$
$= \dfrac{{2\sin 3x\left( {1 - 2{{\sin }^2}x - 1} \right)}}{{\cos 5x - \cos x}} $
$= \dfrac{{2\sin 3x\left( { - 2{{\sin }^2}x} \right)}}{{\cos 5x - \cos x}} $
In the denominator we simplify the expression using the identity $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$:
$= \dfrac{{2\sin 3x\left( { - 2{{\sin }^2}x} \right)}}{{\left( { - 2\sin \left( {\dfrac{{5x + x}}{2}} \right)\sin \left( {\dfrac{{5x - x}}{2}} \right)} \right)}} $
$= \dfrac{{2\sin 3x\left( { - 2{{\sin }^2}x} \right)}}{{\left( { - 2\sin \left( {\dfrac{{6x}}{2}} \right)\sin \left( {\dfrac{{4x}}{2}} \right)} \right)}} $
$= \dfrac{{2\sin 3x\left( { - 2{{\sin }^2}x} \right)}}{{\left( { - 2\sin 3x\left( {\sin 2x} \right)} \right)}} $
$= \dfrac{{2 \times ( - 2)\sin 3x \times {{\sin }^2}x}}{{\left( { - 2\sin 3x \times \left( {2\sin x\cos x} \right)} \right)}} $
$= \dfrac{{2 \times ( - 2)\sin 3x \times \sin x \times \sin x}}{{2 \times \left( { - 2\sin 3x \times \left( {\sin x\cos x} \right)} \right)}} $
on cancelling the common terms from the numerator and denominator we get:
$= \dfrac{{\sin x}}{{\cos x}} $
$= \tan x $
Hence it is proved that the value of \[\dfrac{{{\text{sin5x + sinx - 2sin3x}}}}{{{\text{cos5x - cosx}}}} = \tan x\].

Note: This problem is solved by applications of compound angles and simple multiplication, addition, division. Generally, trigonometric problems seem to be difficult due to the lengthy equations but it is the simpler topic in mathematics until you have a grip on using the appropriate formula at the appropriate time.