Question

Pls answer this..plsss

Answer 1

p4=cos^4a+sin^4a=1-2sin^2acos^a
p6=cos^6a+sin^6a=1-(3cos^4asin^2a+3sin^4acos^2a)=1-3cos^2asin^2a(1)

p6=1-3cos^2asin^2a
p4=1-2sin^2acos^2a
we have to find 3p4-2p6
=3(1-2sin^2acos^2a)-2(1-3cos^2asin^2a)
=3-2=1





I have used 2 formulas
(cos^2a +Sin^2a) ^ 2 =cos^4a+sin^4a+2sin^2acos^2a..
1=(cos^4a+sin^4a)+2sin^2cos^2a....in p4
and
(cos^2a+sin^2a)^3=cos^6a+sin^6a+3cos^2asin^2a(sin^2a+cos^2a)
and sin^2a+cos^2a=1
in)p6



final ans=1