91Ó°ÊÓ

We are given the equation $\frac{\cos \mathrm{θ}-\cos \left(3\mathrm{θ}\right)}{\sin \left(3\mathrm{θ}\right)-\sin \mathrm{θ}}=\tan \left(2\mathrm{θ}\right)$and we need to establish it as an identity.

We first simplify the L.H.S. using the sum to product formula.

The standard formulae required are

$$\begin{gathered} \sin \mathrm{Î\pm }-\sin \mathrm{β}=2\sin \left(\frac{\mathrm{Î\pm }-\mathrm{β}}{2}\right)\cos \left(\frac{\mathrm{Î\pm }+\mathrm{β}}{2}\right) \\ \cos \mathrm{Î\pm }-\cos \mathrm{β}=-2\sin \left(\frac{\mathrm{Î\pm }+\mathrm{β}}{2}\right)\sin \left(\frac{\mathrm{Î\pm }-\mathrm{β}}{2}\right) \end{gathered}$$

Comparing it with the given expression in the numerator we get, $\mathrm{Î\pm }=\mathrm{θ},\mathrm{β}=3\mathrm{θ}$.

Using the above formula we get,

role="math" localid="1646469979777" $$\begin{gathered} \frac{\cos \mathrm{θ}-\cos \left(3\mathrm{θ}\right)}{\sin \left(3\mathrm{θ}\right)-\sin \mathrm{θ}} \\ =\frac{-2\sin \left(\frac{\mathrm{θ}+3\mathrm{θ}}{2}\right)\sin \left(\frac{\mathrm{θ}-3\mathrm{θ}}{2}\right)}{2\sin \left(\frac{3\mathrm{θ}-\mathrm{θ}}{2}\right)\cos \left(\frac{3\mathrm{θ}+\mathrm{θ}}{2}\right)} \\ =\frac{-2\sin \left(\frac{4\mathrm{θ}}{2}\right)\sin \left(-\frac{2\mathrm{θ}}{2}\right)}{2\sin \left(\frac{2\mathrm{θ}}{2}\right)\cos \left(\frac{4\mathrm{θ}}{2}\right)} \\ =\frac{-2\sin \left(2\mathrm{θ}\right)\sin \left(-\mathrm{θ}\right)}{2\sin \left(\mathrm{θ}\right)\cos \left(2\mathrm{θ}\right)} \end{gathered}$$

We know, $\sin \left(-\mathrm{θ}\right)=-\sin \mathrm{θ}$.

$$\begin{gathered} ∴\frac{-2\sin \left(2\mathrm{θ}\right)\sin \left(-\mathrm{θ}\right)}{2\sin \left(\mathrm{θ}\right)\cos \left(2\mathrm{θ}\right)} \\ =\frac{-\left(-2\sin \left(2\mathrm{θ}\right)\sin \left(\mathrm{θ}\right)\right)}{2\sin \left(\mathrm{θ}\right)\cos \left(2\mathrm{θ}\right)} \\ =\frac{2\sin \left(2\mathrm{θ}\right)\sin \left(\mathrm{θ}\right)}{2\sin \left(\mathrm{θ}\right)\cos \left(2\mathrm{θ}\right)} \end{gathered}$$

We cancel the common term $2\sin \left(\mathrm{θ}\right)$from both numerator and denominator and we get $\frac{\sin \left(2\mathrm{θ}\right)}{\cos \left(2\mathrm{θ}\right)}$.

Using the identity $\tan \mathrm{Î\pm }=\frac{\sin \mathrm{Î\pm }}{\cos \mathrm{Î\pm }}$we get,

$$\begin{gathered} \frac{\cos \mathrm{θ}-\cos \left(3\mathrm{θ}\right)}{\sin \left(3\mathrm{θ}\right)-\sin \mathrm{θ}} \\ =\frac{\sin \left(2\mathrm{θ}\right)}{\cos \left(2\mathrm{θ}\right)} \\ =\tan \left(2\mathrm{θ}\right) \end{gathered}$$

Thus, we can see that the L.H.S. of the identity is equal to the R.H.S. of the identity.

Therefore, the given equation is established as an identity.