What is $\sin 2x$ formula in terms of $\tan$.

Find the value of $\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}$ in terms of $\mathbf{tan}\mathbf{}\mathbf{x}$:

As we know,

$\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{·}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{·}\mathbf{cos}\mathbf{}\mathbf{x}$

$$\begin{gathered} \therefore \sin 2x=2·\sin x·\cos x\times \frac{\cos x}{\cos x} \\ =2·\left(\frac{\sin x}{\cos x}\right)·{\cos }^{2}x \\ =2·\tan x·\frac{1}{{\sec }^{2}x}\mathbf{[}\mathbf{\because }\frac{\mathbf{sin}\mathbf{}\mathbf{x}}{\mathbf{cos}\mathbf{}\mathbf{x}}\mathbf{=}\mathbf{tan}\mathbf{}\mathbf{x}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{sec}\mathbf{}\mathbf{x}}\mathbf{]} \\ =\frac{2\tan x}{1+{\tan }^{2}x}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{sec}}^{\mathbf{2}}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathbf{}\mathbf{x}\mathbf{]} \end{gathered}$$

Hence, the formula of $\mathbf{sin}\mathbf{2}\mathbf{x}$ is $\frac{\mathbf{2}\mathbf{}\mathbf{tan}\mathbf{}\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathbf{}\mathbf{x}}$