tan^-1a+tan^-1b+tan^-1c=pi,yhen prove that a+b+c=abc
To prove that a + b + c = abc, given tan^(-1)a + tan^(-1)b + tan^(-1)c = π, we will use the property of tangent function.
Since tan^(-1)a + tan^(-1)b + tan^(-1)c = π, we can write tan^(-1)a + tan^(-1)b = π - tan^(-1)c.
By applying the tangent addition formula, we get:
tan(tan^(-1)a + tan^(-1)b) = tan(π - tan^(-1)c)
Using the tangent addition formula, the left-hand side can be simplified as:
(a + b) / (1 - ab) = (tanπ - tan(tan^(-1)c)) / (1 + tanπ * tan(tan^(-1)c))
Note that tan(π) = 0, and tan(tan^(-1)c) = c.
So, we have:
(a + b) / (1 - ab) = -c / (1 + 0 * c)
This simplifies to:
(a + b) / (1 - ab) = -c
Cross-multiplying this equation, we get:
(a + b) = -c(1 - ab)
Expanding the right-hand side:
a + b = -c + abc
Rearranging terms, we get:
a + b + c = abc
Thus, we have proved that a + b + c = abc given tan^(-1)a + tan^(-1)b + tan^(-1)c = π.