# Exact Value of tan 11¼°

How to find the exact value of tan 11¼° using the value of cos 45°?

Solution:

For all values of the angle A we know that, 2 sin$$^{2}$$ $$\frac{A}{2}$$ = 1 - cos A

Again, for all values of the angle A we know that, 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$ = sin A

Now tan 11¼°

= $$\frac{sin 11¼°}{cos 11¼°}$$

= $$\frac{sin 11¼°}{cos 11¼°}$$ × $$\frac{2 sin 11¼°}{2 sin 11¼°}$$

= $$\frac{2 sin^{2} 11¼°}{2 sin 11¼° cos 11¼°}$$

= $$\frac{1 - cos 22½°}{sin 22½°}$$

= $$\frac{1 - \sqrt{\frac{1 + cos 45°}{2}}}{\sqrt{\frac{1 - cos 45°}{2}}}$$

= $$\frac{\sqrt{2} - \sqrt{1 + cos 45°}}{\sqrt{1 - cos 45°}}$$

= $$\frac{\sqrt{2} - \sqrt{1 + \frac{1}{\sqrt{2}}}}{\sqrt{1 - \frac{1}{\sqrt{2}}}}$$

= $$\frac{\sqrt{2} - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}}{\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2}}}}$$

= $$\frac{\sqrt{2\sqrt{2}} - \sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} - 1}}$$

= $$\frac{\sqrt{2\sqrt{2}} - \sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} - 1}}$$ × $$\frac{\sqrt{\sqrt{2} + 1}}{\sqrt{\sqrt{2} + 1}}$$

= $$\frac{\sqrt{2\sqrt{2}}\cdot \sqrt{\sqrt{2} + 1} - \sqrt{(\sqrt{2} + 1)^{2}}}{\sqrt{(\sqrt{2} + 1)(\sqrt{2} - 1)}}$$

= $$\frac{\sqrt{2\sqrt{2}{(\sqrt{2} + 1})}-(\sqrt{2} + 1)}{{\sqrt{2 - 1}}}$$

= $$\sqrt{4 + 2\sqrt{2}} - (\sqrt{2} + 1)$$