Question

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

Answer 1

Hint: We take common trigonometric terms in numerator and denominator. We then find $\cos 2\theta $ in terms of $\sin \theta $ and $\cos \theta $ using the trigonometric identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and ${{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta $. Then substitute in L.H.S of the given result and make necessary calculations to complete the required proof.

Complete step by step answer:
We have been given a trigonometric function which we need to prove that LHS = RHS = \[\tan \theta \].
We have been given that,
LHS = \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }\] - (1)
From the above expression, let us take \[\sin \theta \] common from the numerator and \[\cos \theta \] as common from the denominator.
LHS = \[\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}\]- (2)
We know that,
\[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta \] - (3)
We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
From this above, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]
\[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \]
Let us put \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \] in (3). We get,
\[{{\cos }^{2}}\theta -1+{{\cos }^{2}}\theta =\cos 2\theta \].
\[\therefore \cos 2\theta =2{{\cos }^{2}}\theta -1\] - (4)
Now let us put \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \] in (3). We get,
\[\left( 1-{{\sin }^{2}}\theta \right)-{{\sin }^{2}}\theta =\cos 2\theta \]
\[\therefore \cos 2\theta =1-2{{\sin }^{2}}\theta \] - (5)
Thus we got,
\[\begin{align}
  & \cos 2\theta =1-2{{\sin }^{2}}\theta \\
 & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\
\end{align}\]
Let us put these expressions on (2).
\[\therefore \] LHS = \[\dfrac{\sin \theta \left( 1-2{{\sin }^{2}}\theta \right)}{\cos \theta \left( 2{{\cos }^{2}}\theta -1 \right)}=\dfrac{\sin \theta .\cos 2\theta }{\cos \theta .\cos 2\theta }\]
Now let us cancel out \[\cos 2\theta \] from the numerator and denominator.
\[\therefore \] LHS = \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \].
Thus we proved that, LHS = RHS = \[\tan \theta \].
Hence proved

Note: From the numerator, don’t mistake it to be a formula for \[\sin 3x\]. Always try to simplify a trigonometric function. By simplifying it you will get an idea of how to solve the rest of the function. Remember the formula \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta \], which is the key point for the solution. We should not make mistakes while solving this problem.