# Exact Value of tan 142½°

How to find the exact value of tan 142½° using the value of sin 15° and cos 15°?

Solution:

For all values of the angle A and B we know that, tan (A + B) = $$\frac{tan A + tan B}{1 - tan A tan B}$$,

sin A = 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$

and

cos A = cos$$^{2}$$ $$\frac{A}{2}$$ – sin$$^{2}$$ $$\frac{A}{2}$$

Now tan 142½°

= tan (90 + 52½°)

= - cot 52½°

= $$\frac{-1}{tan 52½°}$$

= $$\frac{-1}{tan (45° + 7½°}$$

= - $$\frac{1 - tan 7½°}{1 + tan 7½°}$$

= - $$\frac{cos 7½° - sin 7½°}{cos 7½° + sin 7½°}$$

=  - $$\frac{(cos 7½° - sin 7½°)(cos 7½° - sin 7½°)}{(cos 7½° + sin 7½°)(cos 7½° - sin 7½°)}$$

= - $$\frac{(cos 7½° - sin 7½°)^{2}}{cos^{2} 7½° - sin^{2} 7½°}$$

= - $$\frac{1 - 2 sin 7½° cos 7½°}{cos^{2} 7½° - sin^{2} 7½°}$$

= - $$\frac{1 - sin 15°}{cos 15°}$$

= - $$\frac{1 - sin (45° - 30°)}{cos (45° - 30°)}$$

= - $$\frac{1 - \frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}}$$

= - $$\frac{2√2 - √3 + 1}{√3 + 1}$$

= - $$\frac{(2√2 - √3 + 1)}{(√3 + 1)}$$  $$\frac{(√3 - 1)}{(√3 - 1)}$$

= - $$\frac{(2√2 - √3 + 1)(√3 - 1)}{3 - 1}$$

= - $$\frac{(2√2(√3 - 1) - (√3 - 1)^{2}}{2}$$

= -[√2(√3 - 1) – (2 - √3)]

= -√6 + √2 + 2 - √3

= 2 + √2 - √3 - √6