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

  • 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 \(\frac{m^2n^2}{2(m^2 - n^2)}\)
jamb 1987mathematics

The correct answer is B. \(\frac{m^2 + n^2}{2mn}\)

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

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

= Hyp

²= Hyp

= Opp

²= Opp

+ Adj

²+ Adj

Hyp

²Hyp

= (m

²= (m

- n

²- n

) + (2mn)

²) + (2mn)

Hyp

²Hyp

= m

\(^4\)= m

- 2m

²- 2m

n

\(^4\)n

- 4m

²- 4m

- n

²- n

Hyp

²Hyp

= m

\(^4\)= m

+ 2m

²+ 2m

+ n

²+ n

n

Hyp

²Hyp

= (m

²= (m

- n

²- n

)

²)

Hyp

²Hyp

= \(\frac{m^2 + n^2}{2mn}\)

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