Question

Please help me | what is cos^6 x = cot^4*csc^2x-cot^4x????

Answer 1

so cot^6 (x) =cot^4 (x) csc^2 (x)-cot^4 (x) then you just manipulate the right side cot^4(x)csc^2(x)-cot^4(x) =cot^6(x) which is true

Explanation:

Answer 2

Write

csc

4

x

as

1

sin

4

x

and

cot

4

x

as

cos

4

x

sin

4

x

⇒

csc

4

x

−

cot

4

x

=

1

sin

4

x

−

cos

4

x

sin

4

x

=

1

−

cos

4

x

sin

4

x

Now recall that

a

2

−

b

2

=

(

a

−

b

)

⋅

(

a

+

b

)

and use this fact with

a

2

being and

b

2

being

cos

4

x

so that

a

is 1 and b is

cos

2

x

So

csc

4

x

−

cot

4

x

=

1

−

cos

4

x

sin

4

x

=

(

1

−

cos

2

x

)

⋅

1

+

cos

2

x

sin

4

x

But

cos

2

x

+

sin

2

x

=

1

so that

1

−

cos

2

x

=

sin

2

x

so

csc

4

x

−

cot

4

x

=

sin

2

x

⋅

1

+

cos

2

x

sin

4

x

=

1

+

cos

2

x

sin

2

x

=

(

1

sin

2

x

)

+

(

cos

2

x

sin

2

x

)

=

c

s

x

2

x

+

cos

2

x

Explanation: