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(B) Given that $x, y, z$ are sides of a right-angled triangle with $z$ as the hypotenuse, by Pythagoras theorem, we have $x^2 + y^2 = z^2$.This implie

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$2$

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(B) Given that $x, y, z$ are sides of a right-angled triangle with $z$ as the hypotenuse, by Pythagoras theorem, we have $x^2 + y^2 = z^2$.This implies $y^2 = z^2 - x^2$, which can be factored as $y^2 = (z-x)(z+x)$.Taking the logarithm with base $y$ on both sides: $\log_y(y^2) = \log_y((z-x)(z+x))$.$2 = \log_y(z-x) + \log_y(z+x)$.Using the change of base formula $\log_a b = \frac{1}{\log_b a}$, we have $\frac{1}{\log_{z-x} y} = \log_y(z-x)$ and $\frac{1}{\log_{z+x} y} = \log_y(z+x)$.Substituting these into the expression: $\frac{1}{\log_{x+z} y} + \frac{1}{\log_{z-x} y} = \log_y(z+x) + \log_y(z-x) = \log_y((z+x)(z-x)) = \log_y(y^2) = 2$.

If $x, y, z$ are the sides of a right-angled triangle, where $z$ is the hypotenuse (largest side), then find the value of $\frac{1}{\log_{x+z} y} + \frac{1}{\log_{z-x} y}$.

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