Q: Prove that :

( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ). .. … . .. (1 + sec 2ⁿΦ)

September 12, 2021

By Brielle

1 thought on “Q: Prove that :<br /><br />( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ). .. … . .. (1 + sec 2ⁿΦ)<br”

1. Answer:

   tan2 θ – (1/cos2 θ) + 1 = 0

   Solution:

   L.H.S = tan2 θ – (1/cos2 θ) + 1

   = tan2 θ – sec2 θ + 1 [since, 1/cos θ = sec θ]

   = tan2 θ – (1 + tan2 θ) +1 [since, sec2 θ = 1 + tan2 θ]

   = tan2 θ – 1 – tan2 θ + 1

   = 0 = R.H.S. Proved

   2. Verify that:

   1/(sin θ + cos θ) + 1/(sin θ – cos θ) = 2 sin θ/(1 – 2 cos2 θ)

   Solution:

   L.H.S = 1/(sin θ + cos θ) + 1/(sin θ – cos θ)

   = [(sin θ – cos θ) + (sin θ + cos θ)]/(sin θ + cos θ)(sin θ – cos θ)

   = [sin θ – cos θ + sin θ + cos θ]/(sin2 θ – cos2 θ)

   = 2 sin θ/[(1 – cos2 θ) – cos2 θ] [since, sin2 θ = 1 – cos2 θ]

   = 2 sin θ/[1 – cos2 θ – cos2 θ]

   = 2 sin θ/[1 – 2 cos2 θ] = R.H.S. Proved

   3. Prove that:

   sec2 θ + csc2 θ = sec2 θ ∙ csc2 θ

   Solution:

   L.H.S. = sec2 θ + csc2 θ

   = 1/cos2 θ + 1/sin2 θ [since, sec θ = 1/cos θ and csc θ = 1/sin θ]

   = (sin2 θ + cos2 θ)/(cos2 θ sin2 θ)

   = 1/cos2 θ ∙ sin2 θ [since, sin2 θ + cos2 θ = 1]

   = 1/cos2 θ ∙ 1/sin2 θ

   = sec2 θ ∙ csc2 θ = R.H.S. Proved

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