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)
$=1$