Prove $\tan 55^{\circ}\tan 65^{\circ}\tan 75^{\circ}=\tan85^{\circ}$ (Method I)

Let's begin with the left side:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)[/MATH]

Using the product to sum identity for the tangent function:

[MATH]\tan(\alpha)\tan(\beta)=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{\cos(\alpha-\beta)-\cos(\alpha+\beta)}[/MATH]

we may write:

[MATH]\tan\left(65^{\circ} \right)\tan\left(55^{\circ} \right)=\frac{\cos\left(10^{\circ} \right)-\cos\left(120^{\circ} \right)}{\cos\left(10^{\circ} \right)+\cos\left(120^{\circ} \right)}[/MATH]

Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$

Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$.

Let:

[MATH]L=\lim_{x\to\infty}\left(\sin\left(\sqrt{x+1} \right)-\sin\left(\sqrt{x} \right) \right)[/MATH]

Application of the sum-to-product identity: