## Trigonometry (11th Edition) Clone

$$\frac{\tan51^\circ}{1-\tan^251^\circ}=\frac{\tan102^\circ}{2}$$
$$\frac{\tan51^\circ}{1-\tan^251^\circ}$$ Recall the Double-Angle Identity for tangent: $$\tan2A=\frac{2\tan A}{1-\tan^2A}$$ So the formula $\frac{\tan51^\circ}{1-\tan^251^\circ}$ lacks a number $2$ in the numerator for the application of the identity. It can be fixed, though. $$\frac{\tan51^\circ}{1-\tan^251^\circ}=\frac{2\times\frac{1}{2}\times\tan51^\circ}{1-\tan^251^\circ}$$ $$\frac{\tan51^\circ}{1-\tan^251^\circ}=\frac{1}{2}\times\frac{2\tan51^\circ}{1-\tan^251^\circ}$$ Now we can apply the above identity for $\frac{2\tan51^\circ}{1-\tan^251^\circ}$ with $A=51^\circ$ $$\frac{\tan51^\circ}{1-\tan^251^\circ}=\frac{1}{2}\times\tan(2\times51^\circ)$$ $$\frac{\tan51^\circ}{1-\tan^251^\circ}=\frac{\tan102^\circ}{2}$$ As it is hard to get an exact value of $\tan102^\circ$, we leave it unchanged, since the question allows for a single trigonometric function value.