if cosec x-sin x=a*a*a and sec x-cos x=b*b*b then pove that a*a*b*b(a*a+b*b)=1?

Answer 1

cosecx-sinx=a^3
(1-sin^2x)/sinx=a^3
cos^2x/sinx=a^3.....(1)
similarly sin^2x/cosx=b^3......(2)
devide (2)by (1)
sin^3x/cos^3x=(b/a)^3
tanx=b/a
now use (2)
tanx*sinx=b^3
sinx=b^2*a
and from (1) cosx=a^2*b
now use sin^2x+cos^2x=1
and put the above values

Answer 2

cosec x-sin x=a*a*a and sec x-cos x=b*b*b
1 1
----- - sin x = a^3 and ------ - cos x = b^3
sin x cos x

(cos x)^2 = a^3 and (sin x)^2 = b^3 ---------------- (1)


Now u want to prove that

a^2 * b^2 (a^2 + b^2) = 1

Take
L . H . S. = a^2 * b^2 (a^2 + b^2)
= a^4 * b^2 + a^2 * b^4
= (sin x)^2 * a * b^2 + (cos x)^2 * a^2 * b ....... from (1)
= (sin x)^2 * a^3 * b^3 + (cos x)^2 * a^3 * b^3
---------- -----------
a^2 * b a * b^2 (making numerator a^3 * b^3)

= (sin x)^2 * (cos x)^2 * [ (sin x)^2 + (cos x)^2 ]
-------------------------------------------------------
a^3 * b^3
= (sin x)^2 * (cos x)^2 * [1]
---------------------------------
(sin x)^2 * (cos x)^2

= 1 = R. H .S

Answer 3

a^3= 1/sin x - sin x so a^3 = cos^2 x / sinx ------> (1)

b^3 = 1/cosx - cosx so b^3= sin^2 x / cosx ------->(2)

(1)/(2) ----> (a/b)^3 = (cosx/sinx)^3 then a/b=cosx/sinx (3)

now a*a*b*b(a*a + b*b) = a^3 * b^3 (a/b + b/a)=

according to (1),(2),(3) ------>
sinx * cos x* ( (sin^2 x + cos^2 x)/(sinx * cosx) )=

sin^ x + cos ^2 x =1

Answer 4

1/sinx sinx=a^3

or

a^3=cos^2 x/ sinx

1/cosx-cosx=b^3
or

b^3=sin^2x/cosx

a^3b^3=sinxcosx

then try to find
a^4 b^2+a^2b^4=1

Answer 5

I'll assume "cosec" is the cosecant (csc).

a^3 = csc(x) - sin(x)
a^3 = 1/sin(x) - sin(x)
a^3 = 1/sin(x) - sin^2(x)/sin(x)
a^3 = (1 - sin^2(x)) / sin(x)
a^3 = (cos^2(x)) / sin(x)
a^3 = cot(x)cos(x)

b^3 = sec(x) - cos(x)
b^3 = 1/cos(x) - cos^2(x)/cos(x)
b^3 = (1 - cos^2(x)) / cos(x)
b^3 = sin^2(x) / cos(x)
b^3 = sin(x)tan(x)

a^3 / b^3 = cot(x)cos(x) / sin(x)tan(x)
a^3 = b^3 * cot^3(x)
So a = b * cot(x)

(a^2)(b^2)(a^2 + b^2) =
(b^2 * cot^2(x)) (b^2) (b^2 cot^2 (x) + b^2) =
b^6 * cot^2(x) (cot^2(x) + 1) =
b^6 * cot^2(x) csc^2(x) =
(b^3)^2 * cot^2(x) csc^2(x)=
(sec(x) - cos(x))^2 * cot^2(x) csc^2(x)=
[(sec(x) - cos(x)) * cot(x) csc(x)]^2=
[sec(x)cot(x)csc(x) - cos(x)cot(x)csc(x)]^2=
[sec(x) (cos(x)/sin(x)) csc(x) - cos(x)(cos(x)/sin(x))csc(x)]^2 =
[ 1 / sin^2(x) - cos^2(x) / sin^2(x)]^2 =
[ (1 - cos^2(x)) / sin^2(x)]^2 =
[ sin^2(x) / sin^2(x)]^2 =
1^2 = 1

Answer 6

a^3= cos^2x/sinx , b^3 = sin^2x/cosx
Now a^2b^2(a^2+b^2)= a^3b^3(a/b+b/a)
= (cos^2x/sinx)(sin^2x/cosx)
(cosx/sinx+sinx/cosx)
= sinx cosx (sin^2x + cos^2x)/sinx cosx
= 1