Question

Please give solutions of this question​

Answer 1

$\sf{Answer}$

Given to Prove :-

sin⁶ θ + cos⁶θ = 1 - 3sin²θ cos²θ

Formulaes Used :-

a³ + b³ = ( a + b ) ( a² - ab + b²)

sin²θ + cos²θ = 1

( a + b)² = a² + 2ab + b²

By using Above Formulaes We can solve

Solution:-

sin⁶θ + cos⁶θ = 1 - 3sin²θ cos²θ

Take LHS

sin⁶θ + cos⁶θ = (sin²θ)³ + (cos²θ)³

By using a³ + b³ = ( a + b ) ( a² - ab + b²)

a = sin²θ

b = cos²θ

Plugging in formula

a³ + b³ = ( a + b ) ( a² - ab + b²)

(sin²θ)³ + (cos²θ)³ = (sin²θ + cos²θ) [ (sin²θ)² - sin²θcos²θ + (cos²θ)²]

-sin²θ cos²θ can be written as -3sin²θcos²θ + 2cos²θsin²θ

(sin²θ)³ + (cos²θ)³ = 1 ((sin²θ)² +(cos²θ)² + 2cos² θ sin²θ - 3sin² θ cos²θ)

(sin²θ)² +(cos²θ)² + 2cos²θsin²θ it is in form of (a + b)²=a² + 2ab + b²

(sin²θ)³ + ( cos²θ)³ = 1 [ (sin²θ + cos²θ)² - 3sin²θcos²]

sin⁶θ + cos⁶θ = 1[ 1 - 3sin²θcos²θ]

sin⁶θ + cos⁶θ = 1 -3sin²θcos²θ

Hence proved

Know more :-

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj