# 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. $1$

Last updated date: 29th Mar 2023

•

Total views: 309k

•

Views today: 8.88k

Answer

Verified

309k+ 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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India