If $\alpha , \beta , \gamma$ be the three angles given by $\cos \alpha=\dfrac{a}{b+c}, \cos \beta =\dfrac{b}{c+a}$ and $\cos \gamma =\dfrac{c}{a+b}$ where $a,b,c$ are the sides of a $\triangle ABC$, then $\tan \dfrac{\alpha}{2} \tan \dfrac{\beta}{2} \tan \dfrac{\gamma}{2}$ is equal to
Using the half-angle formulas and the given cosine definitions, the expression simplifies using the properties of a triangle where cos(alpha) = a/(b+c). This is a standard trigonometric identity derivation in triangle geometry.
Using the half-angle identity for tangent, tan of alpha divided by 2 equals sine alpha divided by the quantity 1 plus cosine alpha. Since the problem gives cosine alpha as a divided by the quantity b plus c, the denominator becomes 1 plus a divided by b plus c, which simplifies to the sum of a, b, and c divided by b plus c. Using the identity sine alpha equals the square root of 1 minus cosine squared alpha, the numerator becomes the square root of the quantity b plus c minus a multiplied by the quantity b plus c plus a, divided by b plus c. Dividing by the denominator leaves tan of alpha divided by 2 equal to the square root of the quantity b plus c minus a divided by the sum of a, b, and c. Applying the same steps to beta and gamma, and multiplying the three terms together, the numerator becomes the square root of the product of the sums of the sides minus twice a particular side. Because the interior angles of the triangle A, B, and C sum to pi, their half angles sum to pi divided by 2, and using the standard triangle half-angle formula r equals the square root of the product of the differences of the semi-perimeter and the sides divided by the semi-perimeter, the entire product simplifies to r divided by s. This expression is identically equal to the product of tan of A divided by 2, tan of B divided by 2, and tan of C divided by 2, giving the result as the product of tan of A divided by 2, tan of B divided by 2, and tan of C divided by 2.