sin^2(45°- theta) + sin^2(45°+ theta) = 1
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Answer:
To Prove :
Sin²(45-θ)+ Sin²(45+θ) = 1
Let A= 45 - θ, B = 45+θ and θ > 0
=> For θ>0, (45 - θ) will be an acute angle.
For an angle in the 1st Quadrant i.e., for A <90° :
Sin A = Cos (90-A)
=> Sin (45 -θ) =Cos (90-(45-θ))
=> Sin (45 -θ) = Cos (45 +θ)
=> Sin²(45-θ) = Cos²(45+θ) -----(1)
Sin²(45-θ) +Sin²(45+θ) = Cos²(45+θ) + Sin²(45+θ)
= Cos²B + Sin²B
= 1
=> Sin²(45-θ) + Sin²(45+θ) =1. Hence Proved.
Answered by
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To prove :-
Identity used :-
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