# If $lmn = 1$, show that

$\dfrac{1}{{1 + l + {m^{ - 1}}}} + \dfrac{1}{{1 + m + {n^{ - 1}}}} + \dfrac{1}{{1 + n + {l^{ - 1}}}} = 1$

Last updated date: 20th Mar 2023

•

Total views: 308.1k

•

Views today: 5.86k

Answer

Verified

308.1k+ views

Hint- Simplify the L.H.S and reduce it to a single term and try to cancel out numerator and denominator which gives unit value.

Given: $lmn = 1$

To prove: $\dfrac{1}{{1 + l + {m^{ - 1}}}} + \dfrac{1}{{1 + m + {n^{ - 1}}}} + \dfrac{1}{{1 + n + {l^{ - 1}}}} = 1$

Taking first term, \[\dfrac{1}{{1 + l + {m^{ - 1}}}}\]

$

= \dfrac{1}{{1 + l + \dfrac{1}{m}}} \\

= \dfrac{m}{{m + lm + 1}} \\

$

Taking second term, \[\dfrac{1}{{1 + m + {n^{ - 1}}}}\]

\[

= \dfrac{1}{{1 + m + \dfrac{1}{n}}} \\

= \dfrac{1}{{1 + m + lm}}{\text{ }}\left\{ {\because lmn = 1 \Rightarrow \dfrac{1}{n} = lm} \right\} \\

\]

Taking third term, \[\dfrac{1}{{1 + n + {l^{ - 1}}}}\]

Multiply and divide by $lm$, we get:

\[

= \dfrac{1}{{1 + n + {l^{ - 1}}}} \times \dfrac{{lm}}{{lm}} \\

= \dfrac{{lm}}{{lm + lmn + {l^{ - 1}}lm}} \\

= \dfrac{{lm}}{{lm + 1 + m}}{\text{ }}\left\{ {\because lmn = 1} \right\} \\

\]

Now, combining all these terms to form the L.H.S, we get:

\[

{\text{L}}{\text{.H}}{\text{.S}} = \dfrac{1}{{1 + l + {m^{ - 1}}}} + \dfrac{1}{{1 + m + {n^{ - 1}}}} + \dfrac{1}{{1 + n + {l^{ - 1}}}} \\

= \dfrac{1}{{1 + l + \dfrac{1}{m}}} + \dfrac{1}{{1 + m + \dfrac{1}{n}}} + \dfrac{1}{{1 + n + \dfrac{1}{l}}} \\

= \dfrac{m}{{m + lm + 1}} + \dfrac{1}{{1 + m + lm}} + \dfrac{{lm}}{{lm + 1 + m}} \\

\]

Since, these have common denominator, hence we can add them directly.

\[

= \dfrac{{m + 1 + lm}}{{m + lm + 1}} \\

= 1 \\

= {\text{R}}{\text{.H}}{\text{.S}} \\

\]

Hence Proved.

Note- Whenever you see equations like these, always try to look for patterns to reduce the fraction and try to make denominators of each term equal in order to add the terms easily.

Given: $lmn = 1$

To prove: $\dfrac{1}{{1 + l + {m^{ - 1}}}} + \dfrac{1}{{1 + m + {n^{ - 1}}}} + \dfrac{1}{{1 + n + {l^{ - 1}}}} = 1$

Taking first term, \[\dfrac{1}{{1 + l + {m^{ - 1}}}}\]

$

= \dfrac{1}{{1 + l + \dfrac{1}{m}}} \\

= \dfrac{m}{{m + lm + 1}} \\

$

Taking second term, \[\dfrac{1}{{1 + m + {n^{ - 1}}}}\]

\[

= \dfrac{1}{{1 + m + \dfrac{1}{n}}} \\

= \dfrac{1}{{1 + m + lm}}{\text{ }}\left\{ {\because lmn = 1 \Rightarrow \dfrac{1}{n} = lm} \right\} \\

\]

Taking third term, \[\dfrac{1}{{1 + n + {l^{ - 1}}}}\]

Multiply and divide by $lm$, we get:

\[

= \dfrac{1}{{1 + n + {l^{ - 1}}}} \times \dfrac{{lm}}{{lm}} \\

= \dfrac{{lm}}{{lm + lmn + {l^{ - 1}}lm}} \\

= \dfrac{{lm}}{{lm + 1 + m}}{\text{ }}\left\{ {\because lmn = 1} \right\} \\

\]

Now, combining all these terms to form the L.H.S, we get:

\[

{\text{L}}{\text{.H}}{\text{.S}} = \dfrac{1}{{1 + l + {m^{ - 1}}}} + \dfrac{1}{{1 + m + {n^{ - 1}}}} + \dfrac{1}{{1 + n + {l^{ - 1}}}} \\

= \dfrac{1}{{1 + l + \dfrac{1}{m}}} + \dfrac{1}{{1 + m + \dfrac{1}{n}}} + \dfrac{1}{{1 + n + \dfrac{1}{l}}} \\

= \dfrac{m}{{m + lm + 1}} + \dfrac{1}{{1 + m + lm}} + \dfrac{{lm}}{{lm + 1 + m}} \\

\]

Since, these have common denominator, hence we can add them directly.

\[

= \dfrac{{m + 1 + lm}}{{m + lm + 1}} \\

= 1 \\

= {\text{R}}{\text{.H}}{\text{.S}} \\

\]

Hence Proved.

Note- Whenever you see equations like these, always try to look for patterns to reduce the fraction and try to make denominators of each term equal in order to add the terms easily.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE