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If sin θ = $(m^{2} - n^{2}) / (m^{2} + n^{2})$ , with $m > n > 0$ and θ acute, then tan θ equals

Competition practiceTrigonometry

A $(m^{2} - n^{2}) / (2 mn)$
B $2 mn / (m^{2} - n^{2})$
C $(m^{2} + n^{2}) / (2 mn)$
D $2 mn / (m^{2} + n^{2})$

Step-by-Step Solution

Adjacent side = √

[(m^{2} + n^{2})^{2} - (m^{2} - n^{2})^{2}]

= √

(4 m^{2} n^{2}) = 2 mn

, so tan θ =

(m^{2} - n^{2}) / (2 mn)

.

Answer: A)

(m^{2} - n^{2}) / (2 mn)

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