$\frac{{\sqrt {{a^2} + {b^2}} }}{{\sin \alpha \,\,\cos \alpha }}\,\,\,\left[ {\frac{a}{{\sqrt {{a^2} + {b^2}} }}\,\,\cos \alpha \, - \,\frac{b}{{\sqrt {{a^2} + {b^2}} }}\,\sin \alpha } \right]$
Now $sin3\alpha =$ $\frac{a}{{\sqrt {{a^2} + {b^2}} }}$ gives
$ \Rightarrow \,\,\sqrt {{a^2} + {b^2}} \,\,\left[ {\frac{{\sin 3\alpha \,\cos \alpha \,\, - \,\,\cos 3\alpha \,\,\sin \alpha }}{{\sin \alpha \,\,\cos \alpha }}} \right] = 2\sqrt {{a^2}\, + \,{b^2}} $
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