PROVE: cos (6 theta)- cos( 4 theta)/sin(6 theta) + sin(4 theta) = - tan ( theta)
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$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\cos \left(6\mathrm{\theta }\right)-\cos \left(4\mathrm{\theta }\right)}{\sin \left(6\mathrm{\theta }\right)+\sin \left(4\mathrm{\theta }\right)} \\ =\frac{\cos \left(4\mathrm{\theta }+2\mathrm{\theta }\right)-\cos \left(4\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }+2\mathrm{\theta }\right)+\sin \left(4\mathrm{\theta }\right)} \\ =\frac{\cos \left(4\mathrm{\theta }\right).\cos \left(2\mathrm{\theta }\right)-\sin \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)-\cos \left(4\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }\right).\cos \left(2\mathrm{\theta }\right)+\cos \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)+\sin \left(4\mathrm{\theta }\right)} \\ =\frac{\cos \left(4\mathrm{\theta }\right)\left\{\cos \left(2\mathrm{\theta }\right)-1\right\}-\sin \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }\right)\left\{1+\cos \left(2\mathrm{\theta }\right)\right\}+\cos \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)} \\ =-\frac{\cos \left(4\mathrm{\theta }\right)\left\{1-\cos \left(2\mathrm{\theta }\right)\right\}+\sin \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }\right)\left\{1+\cos \left(2\mathrm{\theta }\right)\right\}+\cos \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)} \\ =-\frac{\cos \left(4\mathrm{\theta }\right)\left\{1-1+2{\sin }^2\mathrm{\theta }\right\}+\sin \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }\right)\left\{1+2{\cos }^2\mathrm{\theta }-1\right\}+\cos \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)} \\ =-\frac{\cos \left(4\mathrm{\theta }\right)\left\{2{\sin }^2\mathrm{\theta }\right\}+\sin \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)}{\sin \left(4\mathrm{\theta }\right)\left\{2{\cos }^2\mathrm{\theta }\right\}+\cos \left(4\mathrm{\theta }\right).\sin \left(2\mathrm{\theta }\right)} \\ =-\frac{\cos \left(4\mathrm{\theta }\right)\left\{2{\sin }^2\mathrm{\theta }\right\}+\sin \left(4\mathrm{\theta }\right).\left\{2\sin \left(\mathrm{\theta }\right).\cos \left(\mathrm{\theta }\right)\right\}}{\sin \left(4\mathrm{\theta }\right)\left\{2{\cos }^2\mathrm{\theta }\right\}+\cos \left(4\mathrm{\theta }\right).\left\{2\sin \left(\mathrm{\theta }\right).\cos \left(\mathrm{\theta }\right)\right\}} \\ =-\frac{2\sin \left(\mathrm{\theta }\right)}{2\cos \left(\mathrm{\theta }\right)}\left[\frac{\cos \left(4\mathrm{\theta }\right).\mathrm{sin\theta }+\sin \left(\mathrm{\theta }\right).\sin \left(4\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)\sin \left(4\mathrm{\theta }\right)+\cos \left(4\mathrm{\theta }\right).\sin \left(\mathrm{\theta }\right)}\right] \\ =-\frac{2\sin \left(\mathrm{\theta }\right)}{2\cos \left(\mathrm{\theta }\right)}\left[\frac{\cos \left(4\mathrm{\theta }\right).\mathrm{sin\theta }+\sin \left(\mathrm{\theta }\right).\sin \left(4\mathrm{\theta }\right)}{\cos \left(4\mathrm{\theta }\right).\mathrm{sin\theta }+\sin \left(\mathrm{\theta }\right).\sin \left(4\mathrm{\theta }\right)}\right] \\ \mathbf{=}\mathbf{-}\mathbf{tan}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{R}\mathbf{.}\mathbf{H}\mathbf{.}\mathbf{S}\mathbf{.}\mathbf{ }\left(\mathbf{Hence}\mathbf{ }\mathbf{Proved}\right) \end{gathered}$$

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