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





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11 and 12 Grade Math

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