Question

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cos4a*cosa+sin3a*sin2a/cos2a​

Answer 1

$\displaystyle\bf\\\frac{Cos4\alpha Cos\alpha +Sin3\alpha Sin2\alpha }{Cos2\alpha } =\\\\\\=\frac{\dfrac{1}{2} \Big[Cos(4\alpha -\alpha )+Cos(4\alpha +\alpha )\Big]+\dfrac{1}{2} \Big[Cos(3\alpha -2\alpha )-Cos(3\alpha +2\alpha) \Big]}{Cos2\alpha } =\\\\\\=\frac{Cos3\alpha +Cos5\alpha +Cos\alpha -Cos5\alpha }{2Cos2\alpha } =\frac{Cos3\alpha +Cos\alpha }{2Cos2\alpha } =$

$\displaystyle\bf\\=\frac{2Cos\dfrac{3\alpha +\alpha }{2} Cos\dfrac{3\alpha -\alpha }{2}}{2Cos2\alpha } =\frac{Cos2\alpha Cos\alpha }{Cos2\alpha } =Cos\alpha$