When 0° < θ < 90°, solve the following equation:
sec2 θ - 2 tan θ = 0
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Given,
On Solving,
⇒ 1 + tan2 θ - 2 tan θ = 0
⇒ tan2 θ - 2 tan θ + 1 = 0
⇒ (tan θ - 1)2 = 0
⇒ tan θ - 1 = 0
⇒ tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°.
Hence, the value of θ = 45°.
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If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m2 - n2)2 = 16 mn.
2 cos2 θ + sin θ - 2 = 0
3 cos θ = 2 sin2 θ
tan2 θ = 3 (sec θ - 1).