102.

Read the following information and answer the two items that follow :Let $\frac{\tan \left(3\mathrm{A}\right)}{\tan \left(\mathrm{A}\right)} = \mathrm{K},$ where tan A ≠ 0 and $\mathrm{K} = \frac{1}{3}$

What is tan²(A) equal to ? 

• $\frac{\mathrm{K} + 1}{3\mathrm{K} + 1}$

• $\frac{\mathrm{K} - 1}{3\mathrm{K} - 1}$

• $\frac{3\mathrm{K} - 1}{\mathrm{K} - 1}$

• $\frac{\mathrm{k} + 3}{3\mathrm{k} - 1}$

$$\begin{gathered} \mathrm{We} \mathrm{have}, \frac{\tan \left(3\mathrm{A}\right)}{\tan \left(\mathrm{A}\right)} = \mathrm{K} \\ \Rightarrow \frac{\sin \left(3\mathrm{A}\right) . \cos \left(\mathrm{A}\right)}{\cos \left(3\mathrm{A}\right) . \sin \left(\mathrm{A}\right)} = \mathrm{K} \\ \Rightarrow \frac{\sin \left(3\mathrm{A}\right) . \cos \left(\mathrm{A}\right) + \cos \left(3\mathrm{A}\right) . \sin \left(\mathrm{A}\right)}{\sin \left(3\mathrm{A}\right) . \cos \left(\mathrm{A}\right) - \cos \left(3\mathrm{A}\right) . \sin \left(\mathrm{A}\right)} = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow \frac{\sin \left(3\mathrm{A} + \mathrm{A}\right)}{\sin \left(3\mathrm{A} - \mathrm{A}\right)} = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow \frac{\sin \left(4\mathrm{A}\right)}{\sin \left(2\mathrm{A}\right)} = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow 2\cos \left(2\mathrm{A}\right) = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow 2\left(1 - 2{\sin }^2\left(\mathrm{A}\right)\right) = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow 2 - 4{\sin }^2\left(\mathrm{A}\right) = \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow 4{\sin }^2\left(\mathrm{A}\right) = 2 - \frac{\mathrm{K} + 1}{\mathrm{K} - 1} \\ \Rightarrow 4{\sin }^2\left(\mathrm{A}\right) = \frac{2\mathrm{K} - 2 - \mathrm{K} - 1}{\mathrm{K} + 1} = \frac{\mathrm{K} - 3}{\mathrm{K} - 1} \\ \Rightarrow {\sin }^2\left(\mathrm{A}\right) =\frac{1}{4}\frac{\mathrm{K} - 3}{\mathrm{K} - 1} \\ \mathrm{and} {\cos }^2\left(\mathrm{A}\right) = 1 - \frac{1}{4}\left(\frac{\mathrm{K} - 3}{\mathrm{K} - 1}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow {\cos }^2\left(\mathrm{A}\right) = \frac{4\mathrm{K} - 4 - \mathrm{K} + 3}{\mathrm{K} - 1} = \frac{3\mathrm{K} - 1}{\mathrm{K} - 1} \\ \Rightarrow {\tan }^2\left(\mathrm{A}\right) = \frac{{\sin }^2\left(\mathrm{A}\right) }{{\cos }^2\left(\mathrm{A}\right) } = \frac{\left(\frac{\mathrm{K} - 3}{\mathrm{K} - 1}\right)}{\left(\frac{3\mathrm{K} - 1}{\mathrm{K} - 1}\right)} = \frac{\mathrm{k} - 3}{3\mathrm{k} - 1} \end{gathered}$$