Question

PT cos^4 theta - cos^2 theta = sin^4 theta - sin^2 theta.​

Answer 1

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

Answer 2

Answer:

prove that cos^4θ-cos^2θ=sin^4θ-sin^2θ _ (1)

we know that,

cos2θ=cos^θ-sin^θ

from equation number (1)                                w.k.t       (cos^2θ+sin^2θ)=1

cos^4θ-sin^4θ = cos^2θ-sin^2θ

(cos^4θ-sin^4θ) = 1 x (cos^2θ-sin^2θ)

(cos^4θ-sin^4θ) = (cos^2θ-sin^2θ) x (cos^2θ+sin^2θ)          (a+b) (a-b)=a²-b²

                                      a  -    b                 a     +   b

there fore hence proved that,

cos^4θ-cos^2θ=sin^4θ-sin^2θ