Short & Simple Trig

Geometry Level 2

Given α , β ∈ ( 0 , π / 2 ) \alpha,\beta \in (0,\pi/2) .

If tan ⁡ β = cot ⁡ α − 1 cot ⁡ α + 1 \displaystyle \tan \beta=\frac{\cot \alpha -1}{\cot \alpha +1} ,

find α + β \alpha + \beta .

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

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Danish Ahmed
Jun 23, 2015

tan ⁡ β = 1 − tan ⁡ α 1 + tan ⁡ α \tan\beta=\dfrac{1-\tan\alpha}{1+\tan\alpha}

tan ⁡ α tan ⁡ β + tan ⁡ β = 1 − tan ⁡ α \tan\alpha\tan\beta+\tan\beta=1-\tan\alpha

tan ⁡ α + tan ⁡ β = 1 − tan ⁡ α tan ⁡ β \tan\alpha+\tan\beta=1-\tan\alpha\tan\beta

tan ⁡ α + tan ⁡ β 1 − tan ⁡ α tan ⁡ β = 1 \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=1

tan ⁡ ( α + β ) = 1 \tan(\alpha+\beta)=1

α + β = π 4 \alpha+\beta=\dfrac{\pi}{ 4}

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