# In a right – angled triangle, the sided are $a,b\text{ and }c,$ with $c$ as hypotenuse and

$c-b\ne 1,c+b\ne 1.$ Then the value of $\dfrac{\left( {{\log }_{c+b}}a+{{\log }_{c-b}}a \right)}{\left(

2{{\log }_{c+b}}a\times {{\log }_{c-b}}a \right)}$

A. $2$

B. $-1$

C. $\dfrac{1}{2}$

D. $1$

Answer

Verified

362.7k+ views

Hint: Use ${{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a}$ and basic formula of right angle

triangle.

As we have given that $a,b\text{ and }c$are sides of right – angled triangle with $c$ as

hypotenuse.

By Pythagoras theorem:

${{c}^{2}}={{b}^{2}}+{{a}^{2}}..........\left( i \right)$

The given expression of which we need to find value is:

$\begin{align}

& =\dfrac{{{\log }_{c+b}}a+{{\log }_{c-b}}a}{2{{\log }_{c+b}}a\times {{\log }_{c-b}}a} \\

& =\dfrac{{{\log }_{c+b}}a}{2{{\log }_{c+b}}a\times {{\log }_{c-b}}a}+\dfrac{{{\log

}_{c-b}}a}{2{{\log

}_{c+b}}a\times {{\log }_{c-b}}a} \\

& =\dfrac{1}{2{{\log }_{c-b}}a}+\dfrac{1}{2{{\log }_{c+b}}a} \\

& =\dfrac{1}{2}\left( {{\log }_{a}}\left( c-b \right)+{{\log }_{a}}\left( c+b \right)

\right)\text{

}\because \left( {{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a} \right) \\

& =\dfrac{1}{2}{{\log }_{a}}\left( c-b \right)\left( c+b \right)\text{ }\because \left(

{{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab \right) \\

& =\dfrac{1}{2}{{\log }_{a}}{{c}^{2}}-{{b}^{2}}.................\left( ii \right) \\

\end{align}$

From the equation (i)

${{c}^{2}}={{b}^{2}}+{{a}^{2}}$

Hence, equation (ii) will become

$\begin{align}

& =\dfrac{1}{2}lo{{a}_{a}}{{a}^{2}} \\

& =\dfrac{2}{2}{{\log }_{a}}a\text{ }\left( {{\log }_{x}}{{a}^{n}}=n{{\log }_{x}}a \right) \\

& ={{\log }_{a}}a \\

& =1 \\

\end{align}$

Hence, option D is the correct answer.

Note: (i) One need to remember the basic formula between logarithmic functions as used in

the solution for the flexibility which is an important part of the solution with respect to the

calculation point of view.

(ii) One can change the given expression in following way:

$\begin{align}

& \dfrac{{{\log }_{c+b}}a+{{\log }_{c-b}}a}{2{{\log }_{c+b}}a.{{\log }_{c-b}}a} \\

& =\dfrac{\dfrac{{{\log }_{x}}a}{{{\log }_{x}}\left( c+b \right)}+\dfrac{{{\log }_{x}}a}{{{\log

}_{x}}\left(

c-b \right)}}{\dfrac{2{{\log }_{x}}a}{{{\log }_{x}}\left( c+b \right)}\times \dfrac{{{\log

}_{x}}a}{{{\log

}_{x}}\left( c-b \right)}} \\

\end{align}$

As we have formula ${{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}$ with positive $c$.

And simplifying the above expression will also give the answer, but at the end the same

calculation is required. So no need to change the bases of given logarithm functions. We can

observe the expression and need to solve in the same way as done in the solution.

triangle.

As we have given that $a,b\text{ and }c$are sides of right – angled triangle with $c$ as

hypotenuse.

By Pythagoras theorem:

${{c}^{2}}={{b}^{2}}+{{a}^{2}}..........\left( i \right)$

The given expression of which we need to find value is:

$\begin{align}

& =\dfrac{{{\log }_{c+b}}a+{{\log }_{c-b}}a}{2{{\log }_{c+b}}a\times {{\log }_{c-b}}a} \\

& =\dfrac{{{\log }_{c+b}}a}{2{{\log }_{c+b}}a\times {{\log }_{c-b}}a}+\dfrac{{{\log

}_{c-b}}a}{2{{\log

}_{c+b}}a\times {{\log }_{c-b}}a} \\

& =\dfrac{1}{2{{\log }_{c-b}}a}+\dfrac{1}{2{{\log }_{c+b}}a} \\

& =\dfrac{1}{2}\left( {{\log }_{a}}\left( c-b \right)+{{\log }_{a}}\left( c+b \right)

\right)\text{

}\because \left( {{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a} \right) \\

& =\dfrac{1}{2}{{\log }_{a}}\left( c-b \right)\left( c+b \right)\text{ }\because \left(

{{\log }_{x}}a+{{\log }_{x}}b={{\log }_{x}}ab \right) \\

& =\dfrac{1}{2}{{\log }_{a}}{{c}^{2}}-{{b}^{2}}.................\left( ii \right) \\

\end{align}$

From the equation (i)

${{c}^{2}}={{b}^{2}}+{{a}^{2}}$

Hence, equation (ii) will become

$\begin{align}

& =\dfrac{1}{2}lo{{a}_{a}}{{a}^{2}} \\

& =\dfrac{2}{2}{{\log }_{a}}a\text{ }\left( {{\log }_{x}}{{a}^{n}}=n{{\log }_{x}}a \right) \\

& ={{\log }_{a}}a \\

& =1 \\

\end{align}$

Hence, option D is the correct answer.

Note: (i) One need to remember the basic formula between logarithmic functions as used in

the solution for the flexibility which is an important part of the solution with respect to the

calculation point of view.

(ii) One can change the given expression in following way:

$\begin{align}

& \dfrac{{{\log }_{c+b}}a+{{\log }_{c-b}}a}{2{{\log }_{c+b}}a.{{\log }_{c-b}}a} \\

& =\dfrac{\dfrac{{{\log }_{x}}a}{{{\log }_{x}}\left( c+b \right)}+\dfrac{{{\log }_{x}}a}{{{\log

}_{x}}\left(

c-b \right)}}{\dfrac{2{{\log }_{x}}a}{{{\log }_{x}}\left( c+b \right)}\times \dfrac{{{\log

}_{x}}a}{{{\log

}_{x}}\left( c-b \right)}} \\

\end{align}$

As we have formula ${{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}$ with positive $c$.

And simplifying the above expression will also give the answer, but at the end the same

calculation is required. So no need to change the bases of given logarithm functions. We can

observe the expression and need to solve in the same way as done in the solution.

Last updated date: 22nd Sep 2023

•

Total views: 362.7k

•

Views today: 10.62k