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Answer 1

Answer:

option d) dependent on alpha and beeta

Answer 2

Option (C)

Step-by-step explanation:

Solution :-

Given that

Cos² (α+β) + Cos²(α-β)-Cos 2α Cos 2β

On taking Cos² (α+β)

=> [Cos (α+β)]²

We know that

Cos (A+B) = Cos A Cos B - Sin A Sin B

Cos²(α+β) = (Cos α Cos β - Sin α Sin β)²

=> Cos² α Cos² β + Sin² α Sin² β -

2 Cos α Cos β Sin α Sin β ------(1)

On taking Cos²(α-β)

=> [Cos (α-β)]²

We know that

Cos (A-B) = Cos A Cos B + Sin A Sin B

Cos²(α-β) = (Cos α Cos β +Sin α Sin β)²

=> Cos² α Cos² β + Sin² α Sin² β +

2 Cos α Cos β Sin α Sin β --------(2)

On adding (1)&(2)

=> Cos² (α+β) + Cos²(α-β)

=> Cos² α Cos² β + Sin² α Sin² β - 2 Cos α Cos β Sin α Sin β + Cos² α Cos² β + Sin² α Sin² β + 2 Cos α Cos β Sin α Sin β

=> 2 Cos² α Cos² β +2 Sin² α Sin² β ----(3)

We know that

Cos 2A = Cos² A - Sin² A

Cos 2α = Cos² α- Sin² α ------(4)

Cos 2β = Cos² β- Sin²β ------(5)

Now,

On multiplying (4)&(5)

Cos 2α.Cos 2β

=> (Cos² α- Sin² α)(Cos² β- Sin²β)

=> Cos² α Cos² β - Cos²α Sin²β- Sin² α Cos² β + Sin² α Sin²β -------(6)

Now

On Subtracting (6) from (3)

=> Cos² (α+β) + Cos²(α-β)-Cos 2α Cos 2β

=> (2 Cos² α Cos² β +2 Sin² α Sin² β) - (Cos² α Cos² β - Cos²α Sin²β- Sin² α Cos² β + Sin² α Sin²β)

=> 2 Cos² α Cos² β +2 Sin² α Sin² β - Cos² αCos² β+Cos²α Sin²β+ Sin²αCos² β - Sin² α Sin²β

=> Cos² α Cos² β +Cos²α Sin²β+ Sin² α Cos² β + Sin² α Sin²β

=> Cos²α(Sin²β+Cos² β)+ Sin²α(Cos² β +Sin²β)

=> Cos²α(1)+ sin²α(1)

Since, Sin² A + Cos² A = 1

=> Sin²α + Cos²α

=> 1

It is independent of α and β

Answer:-

The equation Cos² (α+β) + Cos²(α-β)-Cos 2α Cos 2β is an independent of α and β.

Used formulae:-

→ Cos (A+B) = Cos A Cos B - Sin A Sin B

→ Cos (A-B) = Cos A Cos B + Sin A Sin B

→ Cos 2A = Cos² A - Sin² A

→ Sin² A + Cos² A = 1