Question

Cos6a+6cos4a+15cos2a+10/cos5a+5cos3a+10cosa=2cosa

Answer 1

Answer:

(Cos6A + 6Cos4A + 15Cos2A + 10)/(Cos5A + 5Cos3A + 10CosA)  = 2CosA

Step-by-step explanation:

Cos6a+6cos4a+15cos2a+10/cos5a+5cos3a+10cosa=2cosa

Lets solve numerator First

Cos6a+6cos4a+15cos2a+10

= Cos6A + Cos4A + 5Cos4A + 5Cos2A + 10Cos2A + 10

= (Cos6A + Cos4A ) + 5(Cos4A + Cos2A) + 10(Cos2A + 1)

Using CosA + CosB = 2Cos((A+B)/2)Cos((A-B)/2)

& Cos2A + 1 = 2 Cos²A

= 2Cos5ACosA + 5*2Cos3ACosA + 10(2 Cos²A)

= 2CosA(Cos5A + 5Cos3A + 10CosA)

LHS

= 2CosA(Cos5A + 5Cos3A + 10CosA)/ (Cos5A + 5Cos3A + 10CosA)

= 2CosA

= RHS

QED

Proved

(Cos6A + 6Cos4A + 15Cos2A + 10)/(Cos5A + 5Cos3A + 10CosA)  = 2CosA

Answer 2

Answer:

for above given question answer is the 2cos@