Question

Prove that: $\dfrac{{\operatorname{sinA} - 2si{n^3}A}}{{2{{\cos }^3}A - \cos A}} = \tan A$.

Answer 1

Hint: Take the left side of the given equation and try to work with it. Start with taking a common numerator and denominator and use $\dfrac{{\sin A}}{{\cos A}} = \tan A$. Now use ${\sin ^2}A + {\cos ^2}A = 1 \Rightarrow {\sin ^2}A = 1 - {\cos ^2}A$ on the fraction part. Simplify it further to make it equal to the right side of the given equation.

Complete step-by-step answer:
Before moving towards the problem, we should understand the trigonometric ratios better.
Let us take a right-angled triangle with base B, perpendicular P and hypotenuse H.
     

We know by the definition:
$\sin A = \dfrac{{Perpendicular}}{{{\text{Hypotenuse}}}} = \dfrac{P}{H} \Rightarrow {\sin ^2}A = \dfrac{{{P^2}}}{{{H^2}}}$
\[\cos A = \dfrac{{Base}}{{{\text{Hypotenuse}}}} = \dfrac{B}{H} \Rightarrow {\cos ^2}A = \dfrac{{{B^2}}}{{{H^2}}}\]
$\tan A = \dfrac{{Perpendicular}}{{Base}} = \dfrac{P}{B} \Rightarrow {\tan ^2}A = \dfrac{{{P^2}}}{{{B^2}}}$
According to Pythagoras’ theorem, this says that the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides, i.e. ${H^2} = {P^2} + {B^2}$
So, we can say: ${\sin ^2}A + {\cos ^2}A = \dfrac{{{P^2}}}{{{H^2}}} + \dfrac{{{B^2}}}{{{H^2}}} = \dfrac{{{P^2} + {B^2}}}{{{H^2}}} = \dfrac{{{H^2}}}{{{H^2}}} = 1$
And also $\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}} = \dfrac{{\dfrac{{{P^2}}}{{{H^2}}}}}{{\dfrac{{{B^2}}}{{{H^2}}}}} = \dfrac{{{P^2}}}{{{B^2}}} = {\tan ^2}A$
Let’s take the left-hand side of the given equation and try to simplify using trigonometric identities.
Firstly, we can take common $\sin A$ in the numerator and $\cos A$ in the denominator. This will give us:
$ \Rightarrow \dfrac{{\sin A - 2si{n^3}A}}{{2{{\cos }^3}A - \cos A}} = \dfrac{{\sin A\left( {1 - 2{{\sin }^2}A} \right)}}{{\cos A\left( {2{{\cos }^2}A - 1} \right)}}$
Now, as we know that the ratio of sine and cosine is tangent, i.e.$\dfrac{{\sin A}}{{\cos A}} = \tan A$
$ \Rightarrow \dfrac{{\sin A\left( {1 - 2{{\sin }^2}A} \right)}}{{\cos A\left( {2{{\cos }^2}A - 1} \right)}} = \tan A \times \dfrac{{\left( {1 - 2{{\sin }^2}A} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}}$
Also, we have trigonometric identities: ${\sin ^2}A + {\cos ^2}A = 1 \Rightarrow {\sin ^2}A = 1 - {\cos ^2}A$. So, we can substitute this value of ${\sin ^2}A$ in the above relation as:
 \[ \Rightarrow \tan A \times \dfrac{{\left( {1 - 2{{\sin }^2}A} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}} = \tan A \times \dfrac{{\left( {1 - 2\left( {1 - {{\cos }^2}A} \right)} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}}\]
This can be easily simplified by opening up the inner brackets, so we can rewrite it as:
\[ \Rightarrow \tan A \times \dfrac{{\left( {1 - 2\left( {1 - {{\cos }^2}A} \right)} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}} = \tan A \times \dfrac{{\left( {1 - 2 + 2{{\cos }^2}A} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}} = \tan A \times \dfrac{{\left( {2{{\cos }^2}A - 1} \right)}}{{\left( {2{{\cos }^2}A - 1} \right)}}\]
So, we got an equal numerator and denominator. This will give us $1$.
$ \Rightarrow \dfrac{{\operatorname{sinA} - 2si{n^3}A}}{{2{{\cos }^3}A - \cos A}} = \tan A \times 1 = \tan A$
Hence, we proved the left-hand side equal to the right-hand side in the equation.

Note: In the questions where you need to prove in trigonometric ratios, always choose one side on which you can work easily. You should notice that the use of the identities ${\sin ^2}A + {\cos ^2}A = 1$ and $\dfrac{{\sin A}}{{\cos A}} = \tan A$ was a crucial part of the solution. An alternative approach to the problem can be taken by changing ${\cos ^2}A$ in the denominator in form of ${\sin ^2}A$ with the use of the above identities.