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 $\left({\log }_{c+b}a+{\log }_{c-b}a\right)/\left(2{\log }_{c+b}a\times {\log }_{c-b}a\right)$ will be

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

Answer 1

Answer: (D)

Since the triangle is right angle
$\therefore c^2=a^2+b^2$
$\Rightarrow c^2-b^2=a^2$
$\frac{\left[{\log }_{c+b}a+{\log }_{c-b}a\right]}{2{\log }_{c+b}a\times {\log }_{c-b}a}$
$=\frac{\frac{\log a}{\log (c+b)}+\frac{\log a}{\log [c-b]}}{2\frac{\log a}{\log (1+b)}\cdot \frac{\log a}{\log (c-b)}}$
$\left[\because {\log }_{a}b=\frac{\log b}{\log a}\right]$
$=\frac{\log a[\log (c+b)+\log (c-b)]}{2(\log a{)}^2}$
$=\frac{\log (c+b)(c-b)}{2\log a}$
$=\frac{\log \left(c^2-b^2\right)}{\log a^2}$
$=\frac{\log a^2}{\log a^2}$ using Eq. (i)
$=1$