If in a right angled triangle, a and b are the lengths of sides and c is the length of hypotenuse and `c-b ne 1, c+b ne 1`, then show that
`log_(c+b)a+log_(c-b)a=2log_(c+b)a.log_(c-b)a.`

Explanation

We know that in a right angled triangle
`c^(2)=a^(2)+b^(2)`
`c^(2)-b^(2)=a^(2)`……….(i)
LHS = `1/(log_(a)(c+b))+1/(log_(a)(c-b)) = (log_(a)(c-b)+log_(a)(c+b))/(log_(a)(c+b).log_(a)(c-b))`
`=log_(a)(c^(2)-b^(2))/(log_(a)(c+b).log_(a)(c-b)) = (log_(a)a^(2))/(log_(a)(c+b).log_(a)(c-b))` ................(using (i))
`=2/(log_(a)(c+b).log_(a)(c-b))=2log_(c+b)a.log_(c-b)a`=RHS