Question

prove that cos4 theta - cos2 theta = sin4 theta sin2 theta​

Answer 1

CORRECT QUESTION:

Prove that cos⁴ θ - cos² θ = sin⁴ θ - sin² θ

GIVEN:

• cos⁴ θ - cos² θ = sin⁴ θ - sin² θ

TO PROVE:

• cos⁴ θ - cos² θ = sin⁴ - θ sin² θ

EXPLANATION:

METHOD 1:

By taking L.H.S as cos⁴ θ - cos² θ

Take cos² θ as common

cos² θ (cos² θ - 1)

$\boxed{\large{\bold{1-\sin^2 \theta = \cos^2\theta}}}$

$\boxed{\large{\bold{-\sin^2 \theta = \cos^2\theta-1}}}$

cos² θ ( - sin² θ )

(1 - sin² θ )( - sin² θ )

- sin² θ + sin⁴ θ

sin⁴ θ - sin² θ

HENCE PROVED.

METHOD 2:

By taking R.H.S as sin⁴ θ - sin² θ

Take sin² θ as common

sin² θ (sin² θ - 1)

$\boxed{\large{\bold{1-\cos^2 \theta = \sin^2\theta}}}$

$\boxed{\large{\bold{-\cos^2 \theta = \sin^2\theta}-1}}$

sin² θ ( - cos² θ )

(1 - cos² θ )( - cos² θ)

- cos² θ + cos⁴ θ

cos⁴ θ - cos² θ

HENCE PROVED.

Answer 2

Answer:

Prove that cos⁴ θ - cos² θ = sin⁴ θ - sin² θ

GIVEN:

cos⁴ θ - cos² θ = sin⁴ θ - sin² θ

TO PROVE:

cos⁴ θ - cos² θ = sin⁴ - θ sin² θ

EXPLANATION:

METHOD 1:

By taking L.H.S as cos⁴ θ - cos² θ

Take cos² θ as common

cos² θ (cos² θ - 1)

\boxed{\large{\bold{1-\sin^2 \theta = \cos^2\theta}}}

1−sin

2

θ=cos

2

θ

\boxed{\large{\bold{-\sin^2 \theta = \cos^2\theta-1}}}

−sin

2

θ=cos

2

θ−1

cos² θ ( - sin² θ )

(1 - sin² θ )( - sin² θ )

- sin² θ + sin⁴ θ

sin⁴ θ - sin² θ

HENCE PROVED.

METHOD 2:

By taking R.H.S as sin⁴ θ - sin² θ

Take sin² θ as common

sin² θ (sin² θ - 1)

\boxed{\large{\bold{1-\cos^2 \theta = \sin^2\theta}}}

1−cos

2

θ=sin

2

θ

\boxed{\large{\bold{-\cos^2 \theta = \sin^2\theta}-1}}

−cos

2

θ=sin

2

θ−1

sin² θ ( - cos² θ )

(1 - cos² θ )( - cos² θ)

- cos² θ + cos⁴ θ

cos⁴ θ - cos² θ

HENCE PROVED