Period of tan theta (1+sec2theta)(1+sec4theta)(1+sec8 theta) is π /β then β =
Answers
Given : Period of tan theta(1+sec2theta)(1+sec4thta)(1+sec8theta) is π /β
To find : β
Solution:
Tanθ (1 + Sec2θ) (1 + Sec4θ)(1 + Sec8θ)
= (Sinθ/Cosθ)(1 + 1/Cos2θ)(1 + 1/Cos4θ)(1 + 1/Cos8θ)
= (Sinθ/Cosθ)((1 + Cos2θ)/Cos2θ)((1 + Cos4θ)/Cos4θ)((1 + Cos8θ)/Cos8θ)
Using 1 + Cos2θ = 2Cos²θ
= (Sinθ/Cosθ)( 2Cos²θ)/Cos2θ)((1 + Cos4θ)/Cos4θ)((1 + Cos8θ)/Cos8θ)
= (2SinθCosθ / Cos2θ) ((1 + Cos4θ)/Cos4θ)((1 + Cos8θ)/Cos8θ)
Using Sin2θ = 2SinθCosθ
= (Sin2θ / Cos2θ) ((1 + Cos4θ)/Cos4θ)((1 + Cos8θ)/Cos8θ)
= (Sin2θ / Cos2θ) ((2Cos²2θ)/Cos4θ)((1 + Cos8θ)/Cos8θ)
= (2Sin2θCos2θ /Cos4θ)((1 + Cos8θ)/Cos8θ)
= ( Sin4θ/Cos4θ)((1 + Cos8θ)/Cos8θ)
=( Sin4θ/Cos4θ)((2Cos²4θ)/Cos8θ)
= ( 2Sin4θ Cos4θ)/Cos8θ
= Sin8θ/Cos8θ
= Tan8θ
Period of Tan is π
=> Period of Tan8θ = π
=> Period = π/8
π/8 = π /β
=> β = 8
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