If tan θ + cot θ = 5, and θ is an acute angle, find the value of tan² θ + cot² θ.
23
We are given \(\tan\theta + \cot\theta = 5\) and asked for \(\tan^2\theta + \cot^2\theta\). The link is the algebraic identity \((a + b)^2 = a^2 + b^2 + 2ab\).
With \(a = \tan\theta\) and \(b = \cot\theta\): \((\tan\theta + \cot\theta)^2 = \tan^2\theta + \cot^2\theta + 2\tan\theta\cot\theta\).
The key trigonometric fact is that \(\cot\theta = \dfrac{1}{\tan\theta}\), so their product is \(\tan\theta \cdot \cot\theta = 1\).
Hence the middle term is \(2 \times 1 = 2\), giving \((\tan\theta + \cot\theta)^2 = \tan^2\theta + \cot^2\theta + 2\).
Rearrange for the required sum: \(\tan^2\theta + \cot^2\theta = (\tan\theta + \cot\theta)^2 - 2 = 5^2 - 2\).
This equals \(25 - 2 = 23\).
Key concept: squaring a sum of reciprocal-type terms and using \(\tan\theta\cot\theta = 1\) converts a linear condition into the sum of squares. Hence the value is 23.
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