Question

sin2a+sin5a-sina/cos2a+cos5a+cosa=tan2a

Answer 1

Hope this helps.....

Answer 2

"______________

$We \:know \:that ,$

$\blue { sinA - Sin B = 2cos \Big(\frac{A+B}{2}\Big) sin \Big(\frac{A-B}{2}\Big) }$

$\pink { CosA + Cos B = 2cos \Big(\frac{A+B}{2}\Big) Cos \Big(\frac{A-B}{2}\Big) }$

____________”

$LHS \red{= \frac{sin2A+sin5A-sinA}{cos2A+cos 5A + cos A }}$

$= \frac{ sin 2A + 2cos \Big(\frac{5A+A}{2}\Big) sin \Big(\frac{5A-A}{2}\Big)}{cos 2A + 2cos \Big(\frac{5A+A}{2}\Big) Cos \Big(\frac{5A-A}{2}\Big)}$

$=\frac{ sin 2A + 2cos \Big(\frac{6A}{2}\Big) sin \Big(\frac{4A}{2}\Big)}{cos 2A + 2cos \Big(\frac{6A}{2}\Big) Cos \Big(\frac{4A}{2}\Big)}$

$= \frac{sin 2A + 2cos 3A sin2A }{cos 2A + 2 cos 3A cos 2A} \\= \Big(\frac{sin 2A}{cos 2A} \Big) [\frac{1+2cos 3A}{1+2cos 3A}] \\= \frac{sin 2A}{cos 2A } \\ \green {=tan 2A }\\= RHS$

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