Simple one

Geometry Level 2

tan ⁡ a + tan ⁡ b + tan ⁡ c tan ⁡ a tan ⁡ b tan ⁡ c \large \dfrac{\tan a + \tan b + \tan c}{\tan a \tan b \tan c }

Let a , b a,b and c c denote the measure of the angles of a non-right triangle. Compute the expression above.


The answer is 1.

Richard Ryan by richard ryan , 18, Indonesia

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1 solution

tan ⁡ ( a + b + c ) = tan ⁡ ( a ) + tan ⁡ ( b ) + tan ⁡ ( c ) − tan ⁡ ( a ) tan ⁡ ( b ) tan ⁡ ( c ) 1 − tan ⁡ ( a ) tan ⁡ ( b ) − tan ⁡ ( a ) tan ⁡ ( c ) − tan ⁡ ( b ) tan ⁡ ( c ) \tan(a+b+c) = \dfrac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1-\tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)}
In a triangle, a + b + c = π a + b + c = \pi
tan ⁡ ( a + b + c ) = 0 \tan(a+b+c) = 0
tan ⁡ ( a ) + tan ⁡ ( b ) + tan ⁡ ( c ) = tan ⁡ ( a ) tan ⁡ ( b ) tan ⁡ ( c ) \tan(a) + \tan(b) + \tan(c) = \tan(a)\tan(b)\tan(c)
tan ⁡ ( a ) + tan ⁡ ( b ) + tan ⁡ ( c ) tan ⁡ ( a ) tan ⁡ ( b ) tan ⁡ ( c ) = 1 \dfrac{\tan(a) + \tan(b) + \tan(c) }{\tan(a)\tan(b)\tan(c)} = 1


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