Application of inverse trigonometric fúnction

Calculus Level pending

Evaluate tan ⁡ − 1 1 2 + tan ⁡ − 1 1 3 . \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}.

π 4 \frac{\pi }{4} π 2 \frac{\pi }{2} π 6 \frac{\pi }{6} π 3 \frac{\pi }{3}

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Krishna Singh by Krishna Singh , 22, India

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1 solution

Tom Engelsman
Mar 6, 2021

Let x = arctan ⁡ ( 1 / 2 ) + arctan ⁡ ( 1 / 3 ) . x = \arctan(1/2) + \arctan(1/3). Taking the tangent of both sides yields:

tan ⁡ ( x ) = tan ⁡ ( arctan ⁡ ( 1 / 2 ) + arctan ⁡ ( 1 / 3 ) ) = tan ⁡ ( arctan ⁡ ( 1 / 2 ) ) + tan ⁡ ( arctan ⁡ ( 1 / 3 ) ) 1 − tan ⁡ ( arctan ⁡ ( 1 / 2 ) ) tan ⁡ ( arctan ⁡ ( 1 / 3 ) ) = 1 / 2 + 1 / 3 1 − ( 1 / 2 ) ( 1 / 3 ) = 5 / 6 5 / 6 = 1 \tan(x) = \tan(\arctan(1/2)+\arctan(1/3)) = \frac{\tan(\arctan(1/2)) + \tan(\arctan(1/3))}{1-\tan(\arctan(1/2))\tan(\arctan(1/3))} = \frac{1/2+1/3}{1 - (1/2)(1/3)} = \frac{5/6}{5/6} = 1

or x = arctan ⁡ ( 1 ) = π 4 . x = \arctan(1) = \boxed{\frac{\pi}{4}}.

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