Question

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

1

Step-by-step explanation:

Given :

α + β = γ

To find :

cos²α + cos²β + cos²γ - 2cosα cosβ cosγ

Solution :

=> cos²α + cos²β + cos²γ - 2cosα cosβ cosγ

We know that,

$\boxed{\mathrm{sin^2x+cos^2x = 1}}$

=> 1 - sin²α + cos²β + cos²γ - 2cosα cosβ cosγ

=> 1 + (cos²β - sin²α) + cos²γ - 2cosα cosβ cosγ

We know that,

$\boxed{\mathrm{cos^2A - sin^2B = cos(A + B)cos(A-B) }}$

=> 1 + cos(β + α)cos(β - α) + cos²γ - 2cosα cosβ cosγ

As, α + β = γ, substituting in place of γ, we get,

=> 1 + cos(β + α)cos(β - α) + cos²(α + β) - 2cosα cosβ cosγ

=> 1 + cos(β + α)[cos(β - α) + cos(α + β)] - 2cosα cosβ cosγ

We know that,

$\boxed{\mathrm{cosC + cosD = 2cos\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}$

So, using this

=> 1 + cos(β + α)[2cosβcosα] - 2cosα cosβ cosγ

=> 1 + cosγ (2) cosβ cosα - 2cosα cosβ cosγ

=> 1 + 2cosα cosβ cosγ - 2cosα cosβ cosγ

=> 1  + 0

=> 1

❖ Extra information

⟡ Transformation formulae :

$\boxed{\begin{array}{cc} \boxed{\mathrm{sinC + sinD = 2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{sinC - sinD = 2cos\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{cosC + cosD = 2cos\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)}}\\\\\boxed{\mathrm{cosC - cosD = -2sin\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)}}\end{array}}$

Hope it helps!!