Question

Evaluate $\tan \left[2{\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\right]$

• A. $\frac{7}{25}$
• B. $\frac{24}{25}$
• C. $\frac{-24}{7}$
• D. $\frac{8}{3}$

Answer 1

The correct option is C

$\frac{-24}{7}$

Explanation for correct option

Solving the given trigonometric expression

Given, $\tan \left[2{\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\right]$

Let,

$$\begin{gathered} {\cos }^{-1}(3/5)=\mathrm{A} \\ \Rightarrow \mathrm{cosA}=\frac{3}{5}.....\left(1\right) \end{gathered}$$

We know that,

$$\begin{gathered} {\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1 \\ \Rightarrow {\sin }^2\mathrm{A}=1-{\cos }^2\mathrm{A} \end{gathered}$$

Sustituting equation $\left(1\right)$ in the above equation, we get:

$\Rightarrow {\sin }^2\mathrm{A}=1-{\left(\frac{3}{5}\right)}^2$

$\Rightarrow 1-\frac{9}{25}=\frac{16}{25}$

$\therefore \mathrm{sinA}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

We know, $\mathrm{tanA}=\frac{\mathrm{sinA}}{\mathrm{cosA}}$

Implementing $\mathrm{sinA},\mathrm{cosA}$ values in the above equation, we get:

$$\begin{gathered} \Rightarrow \tan A=\frac{{\displaystyle \frac{4}{5}}}{{\displaystyle \frac{3}{5}}} \\ =\frac{4}{3} \end{gathered}$$

Now,

$\tan \left[2{\cos }^{-1}\frac{3}{5}\right]=\tan \left[2A\right]$

$=\frac{2\mathrm{tanA}}{1-{\tan }^2\mathrm{A}}$

$=\frac{2\times \frac{4}{3}}{1-{\left(\frac{4}{3}\right)}^2}$

$=\frac{\left(\frac{8}{3}\right)}{\frac{\left(9-16\right)}{9}}$

$=\frac{8}{3}\times \frac{9}{-7}$

$=(-\frac{24}{7})$

Hence, the correct answer is Option (C).