# if sec x+cos x=2 then find sec^3x(1+sec^3x)+cos^3x(1+cos^3x)

Harsha
19 Points
6 years ago
Sec x+cos x=2 Cos^2(x)+1=2cosxSolve this equation as a quadratic equationCosx = 1Sec^3(x)(1+sec^3(x))+cos^3(x)(1+cos^3(x))1*2+1*2=4
Mohamedfotoh
15 Points
3 years ago
(sec x+cos x)^3=8
Sec^3 x+3Sec^2 x cosx +3Sec x Cos^2 x + cos^3 x= Sec^3 x+Cos^3 x +3 Sec x Cos x (Sec x+Cos x) =8
Sec^3 x+Cos^3 x = 8 – 6=2
(Sec^3 x+cos^3 x)^2=4
Sec^6 x+Cos^6x +2 = 4
Sec^6 x+Cos^6 x =2
then The final anwer is 4
Ram Kushwah
110 Points
3 years ago
sec x + cos x = 2
1/cos x + cos x=2
1+cos² x =2cos x
cos² x – 2*cos x + 1=0
( cosx-1 )²=1
cos x=1
So sec x=1

Now
sec³ x( 1+sec³ x) + cos³ x( 1+cos³ x)
=1 * (1+1) +1*(1+1)
=2+2=4
Sudhir
15 Points
3 years ago
Given : Sec x + Cos x = 2
To find : value of sec^3x(1+sec^3x)+cos^3x(1+cos^3x)

We know that cos x = 1/ sec x ....... (I)
=> Sec x + (1/sec x) = 2    ..from eq. (I)
=> Sec²x +1 = 2 secx
=> Sec²x -2sec x +1 = 0
=> (Sec x-1)² = 0
=> Sec x = 1
Or cos x = 1

sec^3x(1+sec^3x)+cos^3x(1+cos^3x)
= (1)³[1+(1)³] + (1)³[1+(1)³]
= 2 + 2
= 4