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Question
If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`
[Hint: Express `(cos(theta + Φ))/(cos(theta - Φ)) = m/1` and apply Componendo and Dividendo]
Solution
Given that: cos(θ + Φ) = m cos(θ – Φ)
⇒ `(cos(theta + phi))/(cos(theta - phi)) = m/1`
Using componendo and dividendo theorem, we get
`(cos(theta + phi) + cos(theta - phi))/(cos(theta + phi) - cos(theta - phi)) = (m + 1)/(m - 1)`
⇒ `(2cos((theta + phi + theta - phi)/2).cos((theta+ phi - theta + phi)/2))/(-2sin((theta + phi + theta - phi)/2)*sin((theta + phi - theta + phi)/2)) = (m + 1)/(m - 1)`
⇒ `(costheta.cosphi)/(-sintheta.sinphi) = (m + 1)/(m - 1)`
⇒ `- cot theta . cot phi = (m + 1)/(m - 1)`
⇒ `(-cot phi)/(tantheta) = (m + 1)/(m - 1) - (1 + m)/(1 - m)`
⇒ tan θ = `(1 - m)/(1 + m) cot phi`
Hence proved.
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