Question

If a ^ 2 + b ^ 2 = c ^ 2 then the value of 1/(logc + b a) + 1/(logc - b a) =​

Answer 1

Answer:

The answer is 1

Step-by-step explanation:

   $\displaystyle \frac{1}{\log_{c+b}a}+\displaystyle \frac{1}{\log_{c-b}a}$

$=\log_a(c+b)+\log_a(c-b)$, using $\displaystyle \log_ab=\frac{1}{\log_ba}$

$=\log_a[(c+b)(c-b)]\\=\log_a(c^2-b^2)$

$=\log_a(a^2)$, since $a^2+b^2=c^2\Rightarrow a^2=c^2-b^2$

$=2\log _aa$, since $\log_ab^n=n\log_ab$

$=2(1)=2$, since $\log_aa=1$