Question

(sin theta- cos theta)^2​

Answer 1

★ Given that,

• (sin θ - cos θ)²

★ Let,

$\sf\:\implies (\sin\:\theta -\cos\:\theta)^{2}$

• (a - b)² = a² + b² - 2ab

$\sf\:\implies (\sin\:\theta)^{2} + (\cos\:\theta)^{2} - 2\sin\:\theta\cos\:\theta$

$\sf\:\implies \sin^{2}\:\theta + \cos^{2}\:\theta - 2\sin\:\theta\cos\:\theta$

• sin² θ + cos² θ = 1

$\sf\:\implies 1 - 2\sin\:\theta\cos\:\theta$

$\underline{\boxed{\rm{\purple{\therefore (\sin\: \theta - \cos\:\theta)^{2} = 1 - 2\sin\:\theta\cos\:\theta}}}}\:\orange{\bigstar}$

Answer 2

Answer:

Q: (SinΦ - CosΦ)²

=> (SinΦ)² + (CosΦ)² - 2×SinΦ×CosΦ

=> Sin²Φ + Cos²Φ -2SinΦCosΦ

=> 1 - 2SinΦCosΦ ans.

Step-by-step explanation:

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