Given
#32sin^6(theta)=10-15cos(2theta)+bcos(4theta)+acos(6theta)#
we are to prove that
#tan(theta)tan(5theta)=(a+b-5)/(a-b+7)#
#LHS=32sin^6(theta)#
#=4(2sin^2(theta))^3#
#color(red)(["using formula "2sin^2A=1-cos2A]#
#=4(1-cos2(theta))^3#
#=4-12cos2(theta)+12cos^2 2 theta-4cos^3 2theta#
#color(red)(["using formula "4cos^3A=cos3A+cosA])#
#=4-12cos2(theta)+6(1+cos 4 theta)-cos 6theta-3cos2theta#
#=10-15cos2(theta)+6cos 4 theta-cos 6theta#
#color(blue)"After cancelling common terms from both sides"##color(blue)" the given relation becomes as follows."#
#acos 6theta+bcos4theta=-cos6theta+6cos4theta#
#=>(a+1)cos 6theta=(6-b)cos4theta#
#=>(cos4theta)/(cos6theta)=(a+1)/(6-b)#
By dividendo and componendo we get
#=>(cos4theta-cos6theta) /(cos4theta+cos6theta) =((a+1)-(6-b))/((a+1)+(6-b))#
#=>(2sin5thetasintheta) /(2cos5thetacostheta) =(a+1-6+b)/(a+1+6-b)#
#=>tanthetatan5theta =(a+b-5)/(a-b+7)#
Proved
