Let x, y and z be positive real numbers. Suppose x, y and z are the...

We have, $tan\dfrac{X}{2} + tan\dfrac{Z}{2} = \dfrac{2y}{x + y + z} \text{ and } s = x + y + z =$perimeter of a triangle

Apply half-angle formula and $x + y + z = s$ then,

$⇒ \dfrac{∆}{s\left(s - x\right)} + \dfrac{∆}{s\left(s - z\right)} = \dfrac{y}{s} ⇒ ∆ = \left(s - x\right)\left(s - z\right)$

$⇒ {∆}^{2} = s\left(s - x\right)\left(s - y\right)\left(s - z\right) = {\left(s - x\right)}^{2}{\left(s - z\right)}^{2}$

$⇒ {y}^{2} = {x}^{2} + {z}^{2} ⇒ ∠ Y = 90 °$

$⇒ ∠ Y = ∠ X + ∠ Z$

An option \left(B\right)$is correct.</p><p>Now,$\dfrac{x}{y + z} = \dfrac{\sin X}{\sin Y + \sin Z} = \dfrac{\sin X}{1 + \sin \left(\dfrac{\pi}{2} - X\right)} = \dfrac{2sin\dfrac{X}{2}cos\dfrac{X}{2}}{2\cos^{2}\dfrac{X}{2}} = tan\dfrac{X}{2}$</p><p>An Option \left(C\right)$ is correct.