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If (\sec x=2) and (\tan x=\sqrt{3}), what is the value of (\frac{\sec x+\tan x}{\sec x-\tan x})?

Answer and explanation

Correct answer: (7+4\sqrt{3})

Substitute the given trigonometric values directly into the expression. It becomes \\(\\frac{2+\\sqrt3}{2-\\sqrt3}\\). To simplify it, multiply numerator and denominator by the conjugate of the denominator, \\(2+\\sqrt3\\). The denominator becomes \\(2^2-(\\sqrt3)^2=4-3=1\\), while the numerator becomes \\( (2+\\sqrt3)^2=4+4\\sqrt3+3=7+4\\sqrt3\\).

Therefore the value of the expression is \\(7+4\\sqrt3\\), which is option A. The denominator is positive because 2 is greater than \\(\\sqrt3\\), so no sign issue arises. Rationalising is useful here because it removes the radical from the denominator and gives the exact simplified value.

Related tags

SecantTangentRationalisation

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