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Question
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Solution
Consider `(m^2 - n^2) = (tanA + sinA)^2 - (tanA - sinA)^2`
⇒ [tanA + sinA - (tanA - sinA)] [tanA + sinA + (tanA - sinA)]
⇒ [2sinA][2tanA] = 4sinAtanA
Now LHS = `(m^2 - n^2)^2 = (4sinAtanA)^2 = 16sin^2Atan^2A`
Also , RHS = 16mn = 16(tanA + sinA)(tanA - sinA)
⇒ RHS = 16mn = `16(tan^2A - sin^2A) = 16(sin^2A/cos^2A - sin^2A)`
⇒ `16sin^2A((1 - cos^2A)/cos^2A) = 16sin^2A(sin^2A/cos^2A) = 16sin^2Atan^2A`
Thus , `(m^2 - n^2)^2` = 16mn
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