Question:
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)]
Solution:
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