In ∆ABC, if x=tan(B−C2)tanA2, y=tan(C−A2)tanB2, z=tan(A−B2)tanC2, then x+y+z (in terms of x, y, z only) is

x=tan(B−C2)tanA2=(b−cb+c)cot(A2)tan(A2)=b−cb+c1x=b+cb−c=bc∴ 1+x1−x.1+y1−y.1+z1−z=bc.ca.ab=1⇒(1+x)(1+y)(1+z)=(1−x)(1−y)(1−z)⇒2(x+y+z)=−2(xyz)⇒x+y+z=−xyz.

In ∆ABC, if x=tan(B−C2)tanA2, y=tan(C−A2)tanB2, z=tan(A−B2)tanC2, then x+y+z (in terms of x, y, z only) is

x=tan(B−C2)tanA2=(b−cb+c)cot(A2)tan(A2)=b−cb+c1x=b+cb−c=bc∴ 1+x1−x.1+y1−y.1+z1−z=bc.ca.ab=1⇒(1+x)(1+y)(1+z)=(1−x)(1−y)(1−z)⇒2(x+y+z)=−2(xyz)⇒x+y+z=−xyz.

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In ∆ABC, if $x = \tan (\frac{B - C}{2}) \tan \frac{A}{2}, y = \tan (\frac{C - A}{2}) \tan \frac{B}{2}, z = \tan (\frac{A - B}{2}) \tan \frac{C}{2},$ then $x + y + z$ (in terms of x, y, z only) is

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$x y z$

$2 x y z$

– xyz

$\frac{1}{2} x y z$

answer is C.

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Detailed Solution

$\begin{array}{l} x = \tan (\frac{B - C}{2}) \tan \frac{A}{2} = (\frac{b - c}{b + c}) \cot (\frac{A}{2}) \tan (\frac{A}{2}) = \frac{b - c}{b + c} \\ \frac{1}{x} = \frac{b + c}{b - c} = \frac{b}{c} \\ ∴ \frac{1 + x}{1 - x} . \frac{1 + y}{1 - y} . \frac{1 + z}{1 - z} = \frac{b}{c} . \frac{c}{a} . \frac{a}{b} = 1 \Rightarrow (1 + x) (1 + y) (1 + z) = (1 - x) (1 - y) (1 - z) \\ \Rightarrow 2 (x + y + z) = - 2 (x y z) \Rightarrow x + y + z = - x y z . \end{array}$