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Sum

If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`

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#### 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.

Concept: Transformation Formulae

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