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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

11 and 12 Grade Math

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