Given a 2 + b 2 = c 2 , prove that log b+c a + log b-c a = 2log b+c a * log c-b a

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Ritesh Khatri · Answer

log b+c a + log b-c a = 1/(log a b+c) +1/( log a b-c ) =( log a b+c + log a b-c)/(log a b+c* log a b-c ) = log a b 2 -c 2 /(log a b+c* log a b-c ) = log a a 2 /(log a b+c* log a b-c ) =2 /(log a b+c* log a b-c ) = 2log b+c a * log c-b a

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Given a 2 + b 2 = c 2 , prove that log b+c a + log b-c a = 2log b+c a * log c-b a

Given a²+ b²= c², prove that log_b+ca + log_b-ca = 2log_b+ca * log_c-ba

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Given a 2 + b 2 = c 2 , prove that log b+c a + log b-c a = 2log b+c a * log c-b a

Given a2 + b2 = c2, prove that logb+ca + logb-ca = 2logb+ca * logc-ba

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Given a²+ b²= c², prove that log_b+ca + log_b-ca = 2log_b+ca * log_c-ba