#### 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} = a^{2} + 2ab + b^{2})

\[ \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} + 2 = 4\]

\[ \Rightarrow \tan^2 \theta + \frac{1}{\tan^2 \theta} = 4 - 2 = 2\]

Is there an error in this question or solution?

Solution Prave That: If Tan θ + 1 Tan θ = 2 , Then Show that Tan 2 θ + 1 Tan 2 θ = 2 Concept: Application of Trigonometry.