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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°

= $\frac{sin 11¼°}{cos 11¼°}$

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

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

= $\frac{1 - cos 22½°}{sin 22½°}$

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

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

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

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

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

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

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

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

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

11 and 12 Grade Math

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