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]

Given that [MATH]\cos\left(120^{\circ} \right)=-\frac{1}{2}[/MATH] we now have:

[MATH]\tan\left(65^{\circ} \right)\tan\left(55^{\circ} \right)=\frac{\cos\left(10^{\circ} \right)+\frac{1}{2}}{\cos\left(10^{\circ} \right)-\frac{1}{2}}=\frac{2\cos\left(10^{\circ} \right)+1}{2\cos\left(10^{\circ} \right)-1}[/MATH]

Hence, we may write:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{2\cos\left(10^{\circ} \right)+1}{2\cos\left(10^{\circ} \right)-1}\cdot\tan\left(75^{\circ} \right)[/MATH]

Using the identity:

[MATH]\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}[/MATH]

we have:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\left(2\cos\left(10^{\circ} \right)+1 \right)\sin\left(75^{\circ} \right)}{\left(2\cos\left(10^{\circ} \right)-1 \right)\cos\left(75^{\circ} \right)}[/MATH]

Distributing, there results:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{2\sin\left(75^{\circ} \right)\cos\left(10^{\circ} \right)+\sin\left(75^{\circ} \right)}{2\cos\left(75^{\circ} \right)\cos\left(10^{\circ} \right)-\cos\left(75^{\circ} \right)}[/MATH]

Using the product-to-sum identity:

[MATH]2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin( \alpha-\beta)[/MATH]

we have:

[MATH]2\sin\left(75^{\circ} \right)\cos\left(10^{\circ} \right)=\sin\left(85^{\circ} \right)+\sin\left(65^{\circ} \right)[/MATH]

Using the product-to-sum identity:

[MATH]2\cos(\alpha)\cos(\beta)=\cos(\alpha+\beta)+\cos( \alpha-\beta)[/MATH]

we have:

[MATH]2\cos\left(75^{\circ} \right)\cos\left(10^{\circ} \right)=\cos\left(85^{\circ} \right)+\cos\left(65^{\circ} \right)[/MATH]

And now we may state:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+\sin\left(65^{\circ} \right)+\sin\left(75^{\circ} \right)}{\cos\left(85^{\circ} \right)+\cos\left(65^{\circ} \right)-\cos\left(75^{\circ} \right)}[/MATH]

Rearranging, we have:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+\sin\left(75^{\circ} \right)+\sin\left(65^{\circ} \right)}{\cos\left(85^{\circ} \right)-\left(\cos\left(75^{\circ} \right)-\cos\left(65^{\circ} \right) \right)}[/MATH]

Using the sum-to-product identity:

[MATH]\sin(\alpha)+\sin(\beta)=2\sin\left(\frac{\alpha+ \beta}{2} \right)\cos\left(\frac{\alpha-\beta}{2} \right)[/MATH]

we have:

[MATH]\sin\left(75^{\circ} \right)+\sin\left(65^{\circ} \right)=2\sin\left(70^{\circ} \right)\cos\left(5^{\circ} \right)[/MATH]

Using the sum-to-product identity:

[MATH]\cos(\alpha)-\cos(\beta)=-2\sin\left(\frac{\alpha+ \beta}{2} \right)\sin\left(\frac{\alpha-\beta}{2} \right)[/MATH]

we have:

[MATH]-\left(\cos\left(75^{\circ} \right)-\cos\left(65^{\circ} \right) \right)=2\sin\left(70^{\circ} \right)\sin\left(5^{\circ} \right)[/MATH]

Thus, we now have:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\cos\left(5^{\circ} \right)}{\cos\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\sin\left(5^{\circ} \right)}[/MATH]

Using the co-function identities:

[MATH]\cos(\alpha)=\sin\left(90^{\circ}-\alpha \right)[/MATH]

[MATH]\sin(\alpha)=\cos\left(90^{\circ}-\alpha \right)[/MATH]

we have:

[MATH]\cos\left(5^{\circ} \right)=\sin\left(85^{\circ} \right)[/MATH]

[MATH]\sin\left(5^{\circ} \right)=\cos\left(85^{\circ} \right)[/MATH]

and we now may write:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\sin\left(85^{\circ} \right)}{\cos\left(85^{\circ} \right)+2\sin\left(70^{\circ} \right)\cos\left(85^{\circ} \right)}[/MATH]

Factoring, we get:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)\left(1+2\sin\left(70^{\circ} \right) \right)}{\cos\left(85^{\circ} \right)\left(1+2\sin\left(70^{\circ} \right) \right)}[/MATH]

Dividing out common factors, we now have:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\frac{\sin\left(85^{\circ} \right)}{\cos\left(85^{\circ} \right)}[/MATH]

Using the identity:

[MATH]\tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}[/MATH]

we have:

[MATH]\tan\left(55^{\circ} \right)\tan\left(65^{\circ} \right)\tan\left(75^{\circ} \right)=\tan\left(85^{\circ} \right)[/MATH]

Shown as desired.