Question

if 10sin^4x+15 cos^4x=6. Then find 27 cosec^6x+8 sec^6x

Answer 1

given 10sin^4(x)+15cos^4(x)
let sin^2(x)=a
10a^2 +15(1-a)^2=6
10a^2 +15(1+a^2-2a)=6
25a^2 -30a +9=0
a=3/5. sin^2(x)=3/5
cos^2(x)=4/5
cosec^6(x) +8sec^6(x)
(cosec^2(x))^3 +8(sec^2(x))^3
=(5/3)^3 + 8(5/4)^3
= (125/27)+(125/8)
=4735/216

Answer 2

Answer:

Step-by-step explanation:

10sin^4x +15cos^4x

Let sin^2x=a

10a^2+15(1-a)^2=6

10a^2+15(1+a^2-2a)=6

10a^2 +15 +15a^2-30a-6=0

25a^2+30a+9=0

(3-5a)(3-5a)=0

3-5a=0

a=3/5

Sin^2x=3/5       cosec^2x =5/3

Cos^2x= 1-sin^2x

Cos^2x=1-3/5

Cos^2x=2/5      sec^2x=5/2

27cosec^6x+8sec^6x= 27((cosec^2x)^3)+8((sec^2x)^3)

27(5/3)^3+8(5/2)^3

27×125/27+8×125/8

125+125=250