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Question
Prove that:
If \[\tan\theta + \frac{1}{\tan\theta} = 2\], then show that \[\tan^2 \theta + \frac{1}{\tan^2 \theta} = 2\]
Solution
\[\tan\theta + \frac{1}{\tan\theta} = 2\]
Squaring on both sides, we get
\[\left( \tan\theta + \frac{1}{\tan\theta} \right)^2 = 2^2 \]
\[ \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 \times \tan\theta \times \frac{1}{\tan\theta} = 4\] ... (using (a + b)2 = a2 + 2ab + b2)
\[ \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 = 4\]
\[ \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 = 2\]
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