Question

Prove the equation: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

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 \[\left( 1-2{{\sin }^{2}}\theta \right)\] or it can also be written as \[\left( 2{{\cos }^{2}}\theta -1 \right)\] and hence, use the fact that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\].

Complete step by step solution:
In the given 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 the left-hand side of the given equation and try to convert it into the expression of the right-hand side.
So, we have the left-hand side given as,
\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }\]
We will first take $\sin \theta $ common from the numerator and \[\cos \theta \] 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, applying the above identities, we get that the given expression further changes to,
\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta \times \cos 2\theta }{\cos \theta \times \cos 2\theta }\]
It can also be written as,
\[\dfrac{\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }\]
Now, by cancelling \[\cos 2\theta \] from both numerator and denominator, we get, the following expression as,
\[\dfrac{\sin \theta \cos 2\theta }{\cos \theta \cos 2\theta }=\dfrac{\sin \theta }{\cos \theta }\]
We know, an identity that $\tan \theta $ equal to \[\dfrac{\sin \theta }{\cos \theta }\] we can represent the expression \[\dfrac{\sin \theta }{\cos \theta }\] as $\tan \theta $ which equals the expression of the right-hand side.
So, we get \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \].
Hence, it is proved.

Note: We can also use another identity to simplify and prove the result. After simplifying the expression of left-hand side into the form \[\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}\] , we can use the identity \[{{\sin }^{2}}\theta \] equal to \[\left( 1-{{\cos }^{2}}\theta \right)\] and substitute it. Then, we will get the expression as \[\dfrac{\sin \theta \left( 2{{\cos }^{2}}\theta -1 \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}\]. The common terms can be cancelled off and again we will get on simplification, which is the same as right hand side and thus prove the result.