Trigonometry! #62

Geometry Level 3

Find the value of tan ⁡ ( 2 tan ⁡ − 1 ( 1 5 ) − π 4 ) \tan \left( 2\tan^{-1} \left(\frac {1}{5} \right) - \frac {\pi}{4} \right)

Express your answer in the form of − a b -\frac {a}{b} where a a and b b are positive coprime integers, and enter the value of a + b a+b .

This problem is part of the set Trigonometry .


The answer is 24.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
Feb 21, 2015

First, we evaluate tan ⁡ ( 2 tan ⁡ − 1 ( 1 5 ) ) = 2 tan ⁡ ( tan ⁡ − 1 ( 1 5 ) ) 1 − tan ⁡ 2 ( tan ⁡ − 1 ( 1 5 ) ) = 2 5 1 − ( 1 5 ) 2 = 2 5 24 25 = 5 12 \Large{\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right) \\ = \frac{2\tan\left(\tan^{-1}\left(\frac{1}{5}\right)\right)}{1-\tan^{2}\left(\tan^{-1}\left(\frac{1}{5}\right)\right)} \\ = \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}} \\ = \frac{\frac{2}{5}}{\frac{24}{25}} \\ = \frac{5}{12}}

Now, tan ⁡ ( 2 tan ⁡ − 1 ( 1 5 ) − π 4 ) = tan ⁡ ( 2 tan ⁡ − 1 ( 1 5 ) ) − tan ⁡ ( π 4 ) 1 + ( tan ⁡ ( 2 tan ⁡ − 1 ( 1 5 ) ) ) ( tan ⁡ ( π 4 ) ) = 5 12 − 1 1 + ( 5 12 ) = 5 − 12 5 + 12 − 7 17 \Large{\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right) \\ = \frac{\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)-\tan\left(\frac{\pi}{4}\right)}{1+\left(\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)\right)\left(\tan\left(\frac{\pi}{4}\right)\right)} \\ = \frac{\frac{5}{12}-1}{1+\left(\frac{5}{12}\right)} \\ =\frac{5-12}{5+12} \\ \boxed{-\frac{7}{17}}}

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a little help here, could u pls help me understand the second part

ommkar priyadarshi - 4 years, 7 months ago

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