prove that sec(π/4+x)sec(π/4-x)=2sec2x
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Answer:
sec(π/4+x)sec(π/4-x)=2sec2x
Step-by-step explanation:
Prove that sec(π/4+x)sec(π/4-x)=2sec2x
Secθ = 1/Cosθ
LHS
= sec(π/4+x)sec(π/4-x)
= 1/Cos(π/4+x)Cos(π/4-x)
Cos(A+B) = CosACosB - SinASinB
= 1/(Cos(π/4)Cosx - Sin(π/4)Sinx)((Cos(π/4)Cosx + Sin(π/4)Sinx)
=1/(Cos²(π/4)Cos²x - Sin²(π/4)Sin²x))
Cos²(π/4) = Sin²(π/4) = 1/2
= 1/((Cos²(π/4)(Cos²x - Sin²x)
= 2/ (Cos²x - Sin²x)
=2/ (Cosx Cosx - Sinx SInx)
= 2/ (Cos(x + x)
= 2/ Cos(2x)
= 2 Sec2x
= RHS
QED
Proved
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