Question

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

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

Answer 1

The correct option is D

$1$

Explanation for the correct option:

Finding the value of the expression:

As the given triangle is a right-angled triangle, with $c$ as hypotenuse, so by applying Pythagoras Theorem, we get

$$\begin{gathered} c^2=a^2+b^2 \\ \Rightarrow c^2-b^2=a^2...\left(1\right) \end{gathered}$$

$\begin{array}{rcl}\frac{({\log }_{c+b}a+{\log }_{c-b}a)}{\left(2{\log }_{c+b}a\times {\log }_{c-b}a\right)}& =& \frac{{\displaystyle \frac{\log a}{\log \left(c+b\right)}}+{\displaystyle \frac{\log a}{\log \left(c-b\right)}}}{2\times {\displaystyle \frac{\log a}{\log \left(c+b\right)}}\times {\displaystyle \frac{\log a}{\log \left(c-b\right)}}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{a}}\mathbf{b}\mathbf{=}\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{b}}{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{a}}\mathbf{]}\\ & =& \frac{\log a\times \log \left(c+b\right)+\log a\times \log \left(c-b\right)}{2{(\log a)}^2}\\ & =& \frac{\log a\times \log \left(c^2-b^2\right)}{2{(\log a)}^2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{log}\mathbf{a}\mathbf{+}\mathbf{log}\mathbf{b}\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{)}\mathbf{]}\\ & =& \frac{\log \left(c^2-b^2\right)}{2\log a}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}{\mathbf{a}}^{\mathbf{b}}\mathbf{=}\mathbf{2}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{]}\\ & =& \frac{\log a^2}{\log a^2}\\ & =& 1\end{array}$

Hence, option D is correct.