Question

plzz help me very urgent plzz prove that​

Answer 1

Step-by-step explanation:

$\frac{cos2A + 2cos4A + cos6A}{cosA + 2cos3A + cos5A} \\ \frac{2cos4A cos2A + 2cos4A}{2cos3A \: cos2A + 2cos3A} \\ \frac{cos4A}{cos3A} \\ \frac{cos(3A + A)}{cos3A} \\ \frac{cos3A \: cosA - sin3A \: sinA}{cos3A} \\ cosA - sinA \: tan3A \\ \\ Hope \: it \: will \: help \: you.$

Answer 2

Answer:

here three formula are use to solve it,

$( \cos \alpha + \cos\beta ) = 2 \cos( \frac{ \alpha + \beta }{2} ) \cos( \frac{ \alpha - \beta }{2} ) \\ \cos( \alpha + \beta ) = \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$