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Question
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
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Solution
Given , m = a secA + b tanA and n = a tanA + b secA
`m^2 - n^2 = (asecA + btanA)^2 - (atanA + bsecA)^2`
⇒ `a^2sec^2A + b^2tan^2A + 2ab secAtanA - (a^2tan^2A + b^2 sec^2A + 2ab secAtanA)`
⇒ `sec^2A(a^2 - b^2) + tan^2A(b^2 - a^2) = (a^2 - b^2) [sec^2A - tan^2A]`
⇒ `(a^2 - b^2) ["Since" sec^2A - tan^2A = 1]`
Hence , `m^2 - n^2 = a^2 - b^2`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
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