If \(\tan A + \cot A = 4\), what is the value of \(\tan^2 A + \sec^2 A\)?
\(15 + 8\sqrt{3}\)
Let \(\tan A = t\), so \(\cot A = \dfrac{1}{t}\).
Step 1 – Solve for t:
\(t + \dfrac{1}{t} = 4 \implies t^2 - 4t + 1 = 0\)
\(t = \dfrac{4 \pm \sqrt{16 - 4}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\)
Step 2 – Compute tan²A (taking the larger root for the larger answer):
\(\tan^2 A = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}\)
Step 3 – Use the identity \(\sec^2 A = 1 + \tan^2 A\):
\(\sec^2 A = 1 + (7 + 4\sqrt{3}) = 8 + 4\sqrt{3}\)
Step 4 – Add:
\(\tan^2 A + \sec^2 A = (7 + 4\sqrt{3}) + (8 + 4\sqrt{3}) = 15 + 8\sqrt{3}\)
Hence the value is \(15 + 8\sqrt{3}\).
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