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Trigonometric problem (1)

Proof that,  if \alpha + \beta + \gamma = \frac{ \pi }{2}   then tan \alpha \cdot tan \beta + tan \beta \cdot tan \gamma + tan \gamma \cdot tan \alpha = 1 .

Solution:


tan \alpha \cdot tan \beta + tan \beta \cdot tan \gamma + tan \gamma \cdot tan \alpha = tan \beta \cdot ( tan \alpha + tan \gamma ) + tan \alpha\cdot tan \gamma = \frac{ sin \beta }{ cos \beta } \cdot \frac{ sin( \alpha + \gamma ) }{ cos \alpha \cdot cos \gamma } + \frac{ sin \alpha \cdot sin \gamma }{ cos \alpha \cdot cos \gamma } = \frac{ sin \beta + sin \alpha \cdot sin \gamma }{ cos \alpha \cdot cos \gamma } = \frac{ cos( \alpha + \gamma ) + sin \alpha \cdot sin \gamma }{ cos \alpha \cdot cos \gamma } = \frac{ cos \alpha \cdot cos \gamma }{ cos \alpha \cdot cos \gamma } = 1

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Categories: Trigonometry Tags: , ,
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