Question

Prove that $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

Answer 1

Hint: At first, take common $\sin \theta $ from the numerator and $\cos \theta $ from the denominator and after that, use the following identity $\cos 2\theta $ equals to $1-2{{\sin }^{2}}\theta $ or it can also be written as $2{{\cos }^{2}}\theta -1$ and hence, use the fact that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.

Complete step by step answer:
In the question, we have to prove that, the expression $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }$ equals to or can be written as $\tan \theta $
As we are given the equation as,
\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]
So we will first consider left hand side of the given equation and try to convert in the expression of right hand side.
So, we are given,
\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }\]
We will first take $\sin \theta $ common from the numerator and the $\cos \theta $ common from the denominator, so we get:
\[\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}\]
Now here we know an identity which is $\cos 2\theta $ equals to $2{{\cos }^{2}}\theta -1$ and also that $\cos 2\theta $ equals to $1-2{{\sin }^{2}}\theta $
So we can see that the given expression further changes to:
\[\dfrac{\sin \theta \times \cos 2\theta }{\cos \theta \times \cos 2\theta }\]
We can also write it as
\[\dfrac{\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }\]
Now by cancelling $\cos 2\theta $ from both numerator and also from denominator we get the following expression as:
\[\dfrac{\sin \theta }{\cos \theta }\]
We know an identity that, $\tan \theta $ equals to $\dfrac{\sin \theta }{\cos \theta }$ we can represent the expression $\dfrac{\sin \theta }{\cos \theta }$ as $\tan \theta $ which equals to the expression of right hand side.
Hence it is proved.

Note:
After simplifying the expression of left hand side into $\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}$ we can use the identity ${{\sin }^{2}}\theta $ as $1-{{\cos }^{2}}\theta $ and substitute it to write the expression as \[\dfrac{\sin \theta \left( 2{{\cos }^{2}}\theta -1 \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}\Rightarrow \dfrac{\sin \theta }{\cos \theta }\Rightarrow \tan \theta \]