Question

Prove the following: $\dfrac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }=4\cos 2\theta \cos 4\theta $.

Answer 1

Hint: To solve this question, we can convert every term of the left hand side or the LHS in sin and cos. We also know a few relations like, $\sin a\cos b+\cos a\sin b=\sin \left( a+b \right)$ and, $\sin a\cos b-\cos a\sin b=\sin \left( a-b \right)$. By using these relations, we can prove the required expression.

Complete step-by-step answer:

In this question, we have been asked to prove the expression, $\dfrac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }=4\cos 2\theta \cos 4\theta $. To prove this expression, we will first consider the left hand side or the LHS, that is, $\dfrac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }$. We know that $\tan \alpha =\dfrac{\sin \alpha }{\cos \alpha }$. So, we can substitute it in the LHS and then we can write LHS as, $=\dfrac{\dfrac{\sin 5\theta }{\cos 5\theta }+\dfrac{\sin 3\theta }{\cos 3\theta }}{\dfrac{\sin 5\theta }{\cos 5\theta }-\dfrac{\sin 3\theta }{\cos 3\theta }}$

Now, we will take the LCM in both the numerator and denominator. By doing that we get LHS as, $\begin{align}

  & =\dfrac{\dfrac{\left( \sin 5\theta \right)\left( \cos 3\theta \right)+\left( \cos 5\theta \right)\left( \sin 3\theta \right)}{\left( \cos 5\theta \right)\left( \cos 3\theta \right)}}{\dfrac{\left( \sin 5\theta \right)\left( \cos 3\theta \right)-\left( \sin 3\theta \right)\left( \cos 5\theta \right)}{\left( \cos 5\theta \right)\left( \cos 3\theta \right)}} \\

 & \Rightarrow \dfrac{\left[ \left( \sin 5\theta \right)\left( \cos 3\theta \right)+\left( \cos 5\theta \right)\left( \sin 3\theta \right) \right]\left[ \left( \cos 5\theta \right)\left( \cos 3\theta \right) \right]}{\left[ \left( \sin 5\theta \right)\left( \cos 3\theta \right)-\left( \sin 3\theta \right)\left( \cos 5\theta \right) \right]\left[ \left( \cos 5\theta \right)\left( \cos 3\theta \right) \right]} \\

\end{align}$

We know that $\sin a\cos b+\cos a\sin b=\sin \left( a+b \right)$ and, $\sin a\cos b-\cos a\sin b=\sin \left( a-b \right)$. By applying that here in the above expression, we get the LHS as,

$\begin{align}

  & =\dfrac{\sin \left( 5\theta +3\theta \right)}{\sin \left( 5\theta -3\theta \right)} \\

 & =\dfrac{\sin 8\theta }{\sin 2\theta } \\

\end{align}$

We know that $\sin 2\alpha =2\sin \alpha \cos \alpha $. By applying that in the above expression, we get LHS as,

 $\begin{align}

  & =\dfrac{2\sin 4\theta \cos 4\theta }{2\sin \theta \cos \theta } \\

 & =\dfrac{\sin 4\theta \cos 4\theta }{\sin \theta \cos \theta } \\

\end{align}$

We will again apply $\sin 2\alpha =2\sin \alpha \cos \alpha $ in the above expression and get the LHS as, $=\dfrac{2\sin 2\theta \cos 2\theta \cos 4\theta }{\sin \theta \cos \theta }$

Once again, we will apply $\sin 2\alpha =2\sin \alpha \cos \alpha $ in the above expression and get the LHS as, $\begin{align}

  & =\dfrac{2\left( 2\sin \cos \theta \right)\cos 2\theta \cos 4\theta }{\sin \theta \cos \theta } \\

 & =4\cos 2\theta \cos 4\theta \\

\end{align}$

Which is equal to the right hand side or the RHS of the expression given in the question. Hence, we have proved the expression given in the question, that is, $\dfrac{\tan 5\theta +\tan 3\theta }{\tan 5\theta -\tan 3\theta }=4\cos 2\theta \cos 4\theta $.

Note: To solve this question we should know the relation, $\sin 2\alpha =2\sin \alpha \cos \alpha $, which is derived from using $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$, where $x=y=\alpha $. While solving this question, the students can make a mistake with the trigonometric functions, so the students must know the basic trigonometric functions and relations.