How do you solve \[\dfrac{{\sin 8A\cos A - \sin 6A\cos 3A}}{{\cos 2A\cos A - \sin 3A\sin 4A}} = \tan 2A\] ?

Hint: To solve any problem involving trigonometric we need to remember all the identities. Here we take LHS and we simplify it and we show that it is equal to RHS. To solve this we need to know the formula of \[2\sin A\cos B\] , \[2\cos A\sin B\] , \[2\cos A\cos B\] and \[2\sin A\sin B\] . We also know that the tangent is the ratio of sine to cosine function. Using this we can solve the given problem.

Complete step by step solution:
Given,
 \[\dfrac{{\sin 8A\cos A - \sin 6A\cos 3A}}{{\cos 2A\cos A - \sin 3A\sin 4A}} = \tan 2A\]
Now \[LHS = \dfrac{{\sin 8A\cos A - \sin 6A\cos 3A}}{{\cos 2A\cos A - \sin 3A\sin 4A}}\] and \[RHS = \tan 2A\] .
We have the formulas,
 \[
  2\sin A\cos B = \sin (A + B) + \sin (A - B){\text{ }} - - - - (1) \\
  2\cos A\sin B = \sin (A + B) - \sin (A - B){\text{ }} - - - - (2) \\
  2\cos A\cos B = \cos (A + B) + \cos (A - B){\text{ }} - - - - (3) \\
  2\sin A\sin B = \cos (A - B) - \cos (A + B){\text{ }} - - - - (4) \;
 \]
Now we take LHS,
 \[LHS = \dfrac{{\sin 8A\cos A - \sin 6A\cos 3A}}{{\cos 2A\cos A - \sin 3A\sin 4A}}\]
Now multiply the numerator and the denominator by 2, we have
 \[ = \dfrac{{2\sin 8A\cos A - 2\sin 6A\cos 3A}}{{2\cos 2A\cos A - 2\sin 3A\sin 4A}}\]
Now applying the formula we have,
 \[ = \dfrac{{\left( {\sin (8A + A) + \sin (8A - A)} \right) - \left( {\sin (6A + 3A) + \sin (6A - 3A)} \right)}}{{\left( {\cos \left( {2A + A} \right) + \cos \left( {2A - A} \right)} \right) - \left( {\cos \left( {3A - 4A} \right) - \cos \left( {3A + 4A} \right)} \right)}}\]
In the numerator we applied formula (1) and in the denominator we applies formula (3) and (4).
 \[ = \dfrac{{\left( {\sin (9A) + \sin (7A)} \right) - \left( {\sin (9A) + \sin (3A)} \right)}}{{\left( {\cos \left( {3A} \right) + \cos \left( A \right)} \right) - \left( {\cos \left( { - A} \right) - \cos \left( {7A} \right)} \right)}}\]
We know \[\cos \left( { - \theta } \right) = \cos \left( \theta \right)\] .
 \[ = \dfrac{{\sin (9A) + \sin (7A) - \sin (9A) - \sin (3A)}}{{\cos \left( {3A} \right) + \cos \left( A \right) + \cos \left( {7A} \right) - \cos \left( A \right)}}\]
Cancelling the terms we have,
 \[ = \dfrac{{\sin (7A) - \sin (3A)}}{{\cos \left( {3A} \right) + \cos \left( {7A} \right)}}\]
 \[ = \dfrac{{\sin (5A + 2A) - \sin (5A - 2A)}}{{\cos \left( {5A - 2A} \right) + \cos \left( {5A + 2A} \right)}}\]
Again Applying the formula (2) in the numerator and formula (3)
 \[ = \dfrac{{2\cos (5A)\sin (2A)}}{{2\cos (5A)\cos (2A)}}\]
 \[ = \dfrac{{\sin (2A)}}{{\cos (2A)}}\]
By the definition of tangent we have,
 \[ = \tan \left( {2A} \right)\]
 \[ = RHS\] .
Thus we have \[\dfrac{{\sin 8A\cos A - \sin 6A\cos 3A}}{{\cos 2A\cos A - \sin 3A\sin 4A}} = \tan 2A\] .

Note: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions. A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.