Question

If $\tan \left(\theta \right)=t$, then $\tan \left(2\theta \right)+sec\left(2\theta \right)=$

• A. $\frac{1+t}{1-t}$
• B. $\frac{1-t}{1+t}$
• C. $\frac{2t}{1-t}$
• D. $\frac{2t}{1+t}$

Answer 1

The correct option is A

$\frac{1+t}{1-t}$

Explanation for correct option:

Trigonometry identity:

Given: $\tan \left(\theta \right)=t$

$$\begin{gathered} \therefore \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}+\frac{1}{\cos \left(2\theta \right)}\mathbf{}\mathbf{\left[}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{\right]} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}+\frac{1+{\tan }^2\left(\theta \right)}{1-{\tan }^2\left(\theta \right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left[}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{\right]} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{2\tan \left(\theta \right)+1+{\tan }^2\left(\theta \right)}{1-{\tan }^2\left(\theta \right)} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{{\left(1+\tan \left(\theta \right)\right)}^2}{1-{\tan }^2\left(\theta \right)} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{{\left(1+t\right)}^2}{1-t^2}\mathbf{}\mathbf{\left[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{\right]} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{{\left(1+t\right)}^2}{\left(1-t\right)(1+t)} \\ \Rightarrow \tan \left(2\theta \right)+sec\left(2\theta \right)=\frac{\left(1+t\right)}{\left(1-t\right)} \end{gathered}$$

Hence, option A is correct.