Question

In a right-angled triangle, the sides are a, b and c, with c as hypotenuse and $c-b\ne 1,c+b\ne 1$. Then the value of $\frac{(lo{g}_{c+b}a+lo{g}_{c-b}a)}{(2lo{g}_{c+b}a\times lo{g}_{c-b}a)}$

• A. $2$
• B. $-1$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D $1$

Since the triangle is right angle
$\therefore c^2=a^2+b^2\Rightarrow c^2-b^2=a^2\dots \left(i\right)$
Now,
$\frac{[lo{g}_{c+b}a+lo{g}_{c-b}a]}{2lo{g}_{c+b}a\times lo{g}_{c-b}a}=\frac{\frac{loga}{log(c+b)}+\frac{loga}{log[c-b]}}{2\frac{loga}{log(1+b)}.\frac{loga}{log(c-b)}}\left[\begin{array}{c}\because lo{g}_{a}b=\frac{logb}{loga}\end{array}\right]$
$=\frac{loga\left[log\right(c+b)+log(c-b\left)\right]}{2(loga{)}^2}=\frac{log(c+b)(c-b)}{2loga}$
$=\frac{log(c^2-b^2)}{log{a}^2}$
$\frac{log{a}^2}{log{a}^2}$ using Eq. (i)
$=1$