Question

Prove that cot2 theta + cosec2 theta = cosec4 theta – cot4 theta. ​

Answer 1

GIVEN:

• cot² θ + cosec² θ = cosec⁴ θ – cot⁴ θ

TO PROVE:

• cot² θ + cosec² θ = cosec⁴ θ – cot⁴ θ

FORMUALE USED:

$\boxed{\large{\bold{A^2 - B^2 = (A+B)(A-B)}}}$

$\boxed{\large{\bold{cosec^2 \theta - cot^2 \theta = 1 }}}$

PROOF:

METHOD 1:

Take cot² θ + cosec² θ as L.H.S and cosec⁴ θ – cot⁴ θ as R.H.S.

Take the R.H.S

1(cot² θ + cosec² θ)

$\boxed{\large{\bold{cosec^2 \theta - cot^2 \theta = 1 }}}$

(cosec² θ + cot² θ)(cosec² θ – cot² θ)

Apply (A + B)(A - B)  = A² - B²

cosec⁴ θ – cot⁴ θ

L.H.S =  R.H.S

cot² θ + cosec² θ = cosec⁴ θ – cot⁴ θ

HENCE PROVED.

METHOD 2:

Take cot² θ + cosec² θ as L.H.S and cosec⁴ θ – cot⁴ θ as R.H.S.

Take the R.H.S

cosec⁴ θ – cot⁴ θ

Apply A² - B² = (A + B)(A - B)

(cosec² θ + cot² θ)(cosec² θ – cot² θ)

$\boxed{\large{\bold{cosec^2 \theta- cot^2 \theta = 1 }}}$

(cosec² θ – cot² θ) = 1

(cosec² θ + cot² θ) = L.H.S

L.H.S = R.H.S

cot² θ + cosec² θ = cosec⁴ θ – cot⁴ θ

HENCE PROVED.