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Question
If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]
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Solution
tanθ + sinθ = m ......(i)
tanθ – sinθ = n ......(ii)
Adding equations i and ii
2tanθ = m + n ......(iii)
Subtracting equation ii from i
We get,
2sinθ = m – n ......(iv)
Multiplying equations (iii) and (iv)
2sinθ(2tanθ) = (m + n)(m – n)
⇒ 4sinθ tanθ = m2 – n2
Hence,
m2 – n2 = 4sinθ tanθ
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