Q: Prove that :
( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ).. .. .. .. (1 + sec 2ⁿΦ)
Solve it fast :
Answers
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
Step-by-step explanation:
Refer to the above attachment !!
hope it will help u !!
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