Question

give the right answer

Answer 1

Step-by-step explanation:

Sin⁸θ-cos⁸θ

=(sin⁴θ)²-(cos⁴θ)²

=(sin⁴θ+cos⁴θ)(sin⁴θ-cos⁴θ)

={(sin²θ)²+(cos²θ)²}{(sin²θ)²-(cos²θ)²}

={(sin²θ+cos²θ)²-2sin²θcos²θ}{(sin²θ+cos²θ)(sin²θ-cos²θ)}

={(1)²-2sin²θcos²θ}{(1)(sin²θ-cos²θ)} [∵, sin²θ+cos²θ=1]

=(sin²θ-cos²θ)(1-2sin²θcos²θ) (Proved)

Answer 2

$\huge\red Answer$

To Prove

(cos^8∅ - sin^8∅) = (cos²∅ - sin²∅)(1 - 2 sin²∅ cos²∅)

Proving

L.H.S → (cos^8∅ - sin^8∅) = (cos⁴∅)² - (sin⁴∅)²

→ (cos⁴∅ + sin⁴∅)(cos⁴∅ - sin⁴∅)

→ [(cos²∅ + sin²∅)² - 2cos²∅sin²∅][cos²∅ + sin²∅](cos²∅ - sin²∅)

Since cos²∅ + sin²∅ = 1,

→ (1 - 2cos²∅ sin²∅)(cos²∅ - sin²∅) = R.H.S

Hence Proved