In a right-angled triangle,the sides are a, b | vedclass

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(D) Given that $a, b, c$ are sides of a right-angled triangle with $c$ as the hypotenuse,we have $a^2 + b^2 = c^2$,which implies $a^2 = c^2 - b^2 = (c

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$1$

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(D) Given that $a, b, c$ are sides of a right-angled triangle with $c$ as the hypotenuse,we have $a^2 + b^2 = c^2$,which implies $a^2 = c^2 - b^2 = (c+b)(c-b)$.Taking the logarithm on both sides,we get $\log(a^2) = \log((c+b)(c-b)) = \log(c+b) + \log(c-b)$.Now,the given expression is $E = \frac{\log_{c+b} a + \log_{c-b} a}{2 \log_{c+b} a \cdot \log_{c-b} a}$.Using the change of base formula $\log_x y = \frac{\log y}{\log x}$,we have:$E = \frac{\frac{\log a}{\log(c+b)} + \frac{\log a}{\log(c-b)}}{2 \cdot \frac{\log a}{\log(c+b)} \cdot \frac{\log a}{\log(c-b)}}$.Simplifying the numerator: $\frac{\log a (\log(c-b) + \log(c+b))}{\log(c+b) \log(c-b)}$.Simplifying the denominator: $\frac{2 (\log a)^2}{\log(c+b) \log(c-b)}$.Thus,$E = \frac{\log a (\log((c+b)(c-b)))}{2 (\log a)^2} = \frac{\log a \cdot \log(a^2)}{2 (\log a)^2} = \frac{\log a \cdot 2 \log a}{2 (\log a)^2} = \frac{2 (\log a)^2}{2 (\log a)^2} = 1$.

In a right-angled triangle,the sides are $a, b$ and $c$,with $c$ as the hypotenuse,and $c-b \neq 1, c+b \neq 1$. Then the value of $\frac{\log_{c+b} a + \log_{c-b} a}{2 \log_{c+b} a \times \log_{c-b} a}$ is:

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