Final answer:
The equation of the tangent to the curve at the point (1, sqrt(2)) can be found by parameteric differentiation resulting in the line y = x + sqrt(2) - 1. This can be obtained without eliminating the parameter through the derivatives of x and y with respect to the parameter theta. The result is the same when the parameter is eliminated and the derivative is taken directly with respect to x.
Step-by-step explanation:
To find an equation of the tangent to the curve at the given point, we first need to compute the derivative of y with respect to x. Without eliminating the parameter, we do this by finding dy/dθ and dx/dθ and then dividing the two: dy/dx = (dy/dθ)/(dx/dθ).
For x = tan(θ), dx/dθ = sec^2(θ). For y = sec(θ), dy/dθ = sec(θ)tan(θ). At the point (1, sqrt(2)), θ equals π/4, so the slope of the tangent line is sec(π/4)tan(π/4) / sec^2(π/4) = 1. Therefore, the equation of the tangent line is y - sqrt(2) = 1(x - 1), which simplifies to y = x + sqrt(2) - 1.
To eliminate the parameter, we note that x^2 = tan^2(θ) and sec^2(θ) = 1 + tan^2(θ), so y = sqrt(x^2 + 1). The slope is then found using implicit differentiation, which gives us the same result as before.