If tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) find sec\(\theta\)

Mathematics

JAMB 1987

A. \(\frac{m^2 + n^2}{(m^2 - n^2)}\)
B. \(\frac{m^2 + n^2}{2mn}\)
C. \(\frac{mn}{2(m^2 + n^2)}\)
D. \(\frac{m^2n^2}{2(m^2 - n^2)}\)

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Correct Answer: Option B

Explanation

Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\)

\(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem

= Hyp² = Opp² + Adj²

Hyp² = (m² - n²) + (2mn)²

Hyp² = m⁴ - 2m²n⁴ - 4m² - n²

Hyp² = m⁴ + 2m² + n²n

Hyp² = (m² - n²)²

Hyp² = \(\frac{m^2 + n^2}{2mn}\)

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