Question

Prove that \[\dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}} = \dfrac{{\sin 2A}}{{\sin 8A}}\]

Answer 1

Hint:
We will express \[\tan \theta \] as \[\dfrac{{\sin \theta }}{{\cos \theta }}\] in the fraction in the LHS. We will take LCM of the fractions and cancel out common terms. Then, we will express the terms as sum and difference of sine of angles. We will further simplify the equation and obtain the required expression.

Formulas used:
We will use the following formulas:
1) The tangent of an angle \[\theta \] is the ratio of the sine and cosine of that angle:
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2) The Sine of the sum of 2 angles A and B is given by:
\[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
3) The Sine of the difference of 2 angles A and B is given by:
\[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]

Complete step by step solution:
We will use the formula for the tangent of an angle in the numerator and denominator of the fraction on the left-hand side to express it in the form of Sine and Cosine.
Therefore, we get
\[ \Rightarrow \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}} = \dfrac{{\dfrac{{\sin 5A}}{{\cos 5A}} - \dfrac{{\sin 3A}}{{\cos 3A}}}}{{\dfrac{{\sin 5A}}{{\cos 5A}} + \dfrac{{\sin 3A}}{{\cos 3A}}}}\]
We will now take the lowest common multiple (LCM) of the denominators and simplify the fraction. Therefore, we get
\[ \Rightarrow \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}}{\rm{ = }}\dfrac{{\dfrac{{\sin 5A\cos 3A - \sin 3A\cos 5A}}{{\cos 5A\sin 3A}}}}{{\dfrac{{\sin 5A\cos 3A + \sin 3A\cos 5A}}{{\cos 5A\sin 3A}}}}\]
We can see that \[\cos 5A\sin 3A\] is common in the numerator and denominator.
Cancelling out the like terms, we get
\[ \Rightarrow \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}}{\rm{ = }}\dfrac{{\sin 5A\cos 3A - \sin 3A\cos 5A}}{{\sin 5A\cos 3A + \sin 3A\cos 5A}}\]
We will compare the terms in the numerator with the formula \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\] and the terms in the denominator with the formula \[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\].
Expressing the fraction as ratio of Sine of the difference of 2 angles and Sine of the sum of 2 angles, we get
\[ \Rightarrow \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}}{\rm{ = }}\dfrac{{\sin \left( {5A - 3A} \right)}}{{\sin \left( {5A + 3A} \right)}}\]
Subtracting the terms in the brackets in the numerator and adding the terms in the brackets in the denominator, we get
\[ \Rightarrow \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}}{\rm{ = }}\dfrac{{\sin \left( {2A} \right)}}{{\sin \left( {8A} \right)}}\]
We can see that the term on the left-hand side is equal to the term on the right-hand side.

Hence, we have proved that \[\dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}} = \dfrac{{\sin 2A}}{{\sin 8A}}\].

Note:
We can also take the expression on the RHS as our starting point and apply the properties in a reverse order to obtain the LHS.
\[\dfrac{{\sin 2A}}{{\sin 8A}} = \dfrac{{\sin \left( {5A - 3A} \right)}}{{\sin \left( {5A + 3A} \right)}}\]
Now expanding the terms, we get
\[ \Rightarrow \dfrac{{\sin 2A}}{{\sin 8A}}{\rm{ = }}\dfrac{{\sin 5A\cos 3A - \sin 3A\cos 5A}}{{\sin 5A\cos 3A + \sin 3A\cos 5A}}\]
Dividing the numerator and denominator by \[\cos 5A\sin 3A\], we get
\[ \Rightarrow \dfrac{{\sin 2A}}{{\sin 8A}}{\rm{ = }}\dfrac{{\dfrac{{\sin 5A\cos 3A - \sin 3A\cos 5A}}{{\cos 5A\sin 3A}}}}{{\dfrac{{\sin 5A\cos 3A + \sin 3A\cos 5A}}{{\cos 5A\sin 3A}}}}\]
Now dividing the terms, we get
\[ \Rightarrow \dfrac{{\sin 2A}}{{\sin 8A}}{\rm{ = }}\dfrac{{\dfrac{{\sin 5A}}{{\cos 5A}} - \dfrac{{\sin 3A}}{{\cos 3A}}}}{{\dfrac{{\sin 5A}}{{\cos 5A}} + \dfrac{{\sin 3A}}{{\cos 3A}}}}\]
Using the trigonometric identity, \[\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta \], we get
\[ \Rightarrow \dfrac{{\sin 2A}}{{\sin 8A}} = \dfrac{{\tan 5A - \tan 3A}}{{\tan 5A + \tan 3A}}\]
Hence, proved.