If tan θ + sin θ = m and tan θ – sin θ = n, then prove that m2

1.	If tan θ + sin θ = m and tan θ – sin θ = n, then prove that m2 – n2 = 4 sin θ tan θ[Hint: m + n = 2tanθ, m – n = 2 sin θ, then use m2 – n2 = (m + n)(m – n)]
Answer» According to the question, tan θ + sin θ = m …(i) tan θ – sin θ = n …(ii) Adding equation i and ii, 2 tan θ = m + n …(iii) Subtracting equation ii from i, We get, 2sin θ = m – n …(iv) Multiplying equations (iii) and (iv), 2sin θ (2tan θ) = (m + n)(m – n) ⇒ 4 sin θ tan θ = m² – n² Hence, m² – n² = 4 sin θ tan θ

If tan θ + sin θ = m and tan θ – sin θ = n, then prove that m2 – n2 = 4 sin θ tan θ[Hint: m + n = 2tanθ, m – n = 2 sin θ, then use m2 – n2 = (m + n)(m – n)]

Answer»

According to the question,

tan θ + sin θ = m …(i)

tan θ – sin θ = n …(ii)

Adding equation i and ii,

2 tan θ = m + n …(iii)

Subtracting equation ii from i,

We get,

2sin θ = m – n …(iv)

Multiplying equations (iii) and (iv),

2sin θ (2tan θ) = (m + n)(m – n)

⇒ 4 sin θ tan θ = m² – n²

Hence,

m² – n² = 4 sin θ tan θ