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If secθ = $x + 1/ (4 x)$ , then secθ + tanθ equals:

Competition practiceTrigonometry

A $x or 1/ x$
B $2 x or 2/ x$
C $4 x or 1/ (4 x)$
D $2 x or 1/ (2 x)$

Step-by-Step Solution

ta

n^{2}

θ = se

c^{2}

θ −

1 = (x + 1/ (4 x))^{2} - 1 = (x - 1/ (4 x))^{2}

, so tanθ = ±

(x - 1/ (4 x))

. Adding to secθ gives either

2 x

or

2/ (4 x) = 1/ (2 x)

.

Answer: D)

2 x or 1/ (2 x)

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