If a^{2}+b^{2}=c^{2}, then frac{1}{log _{c+a} | vedclass

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(B) Given the expression $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}$.Using the change of base property $\frac{1}{\log _{x} y} = \log _{y} x$,we

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$2$

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(B) Given the expression $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}$.Using the change of base property $\frac{1}{\log _{x} y} = \log _{y} x$,we can rewrite the expression as:$\log _{b}(c+a) + \log _{b}(c-a)$.Using the logarithmic property $\log _{m} n + \log _{m} p = \log _{m} (n \cdot p)$,we get:$\log _{b}((c+a)(c-a)) = \log _{b}(c^{2}-a^{2})$.Since $a^{2}+b^{2}=c^{2}$,it follows that $c^{2}-a^{2}=b^{2}$.Substituting this into the expression:$\log _{b}(b^{2}) = 2 \log _{b} b = 2(1) = 2$.

If $a^{2}+b^{2}=c^{2},$ then $\frac{1}{\log _{c+a} b}+\frac{1}{\log _{c-a} b}=$

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