If 15sin4α+10cos4α=6 then the value of 8cosec6α−27sec6α is
0
1
-1
5
15tan4α+10=61+tan2α2⇒ 9tan4α−12tan2α+4=0⇒ tan2α=2/3 so 8cosec6α−27sec6α=8(1+3/2)3−27(1+2/3)3=0