If 7cos2θ + 5sin2θ − 6 = 0, (0° < θ < 90°), find the value of √((sec θ + tan θ) / (sec θ − tan θ)).
√2 + 1
Start with 7cos2θ + 5sin2θ − 6 = 0 and substitute sin2θ = 1 − cos2θ:
7cos2θ + 5(1 − cos2θ) = 6
2cos2θ + 5 = 6
cos2θ = 1/2, so cos θ = 1/√2 and θ = 45°.
At θ = 45°, sec θ = √2 and tan θ = 1. So (sec θ + tan θ)/(sec θ − tan θ) = (√2 + 1)/(√2 − 1) = (√2 + 1)2/(2 − 1) = 3 + 2√2.
Taking the square root: √(3 + 2√2) = √2 + 1.
Hence, the correct answer is √2 + 1.
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