MOHAN   9 years AGO

GIVEN [1 / 1+a^(n-m)] + [1 / 1+a^(m-n)]

=[1/1+(a^n/a^m)]+[1/1+(a^m/a^n)]

=[1/(a^m+a^n/a^m)]+[1/(a^n+a^m/a^n)]

=[a^m/(a^m+a^n)]+[a^n/(a^n+a^m)]

=[a^m+a^n]/[a^m+a^n]

=1
so answer is 2

Like? Yes (30) | No (15)  

manideep   9 years AGO

->we know that a^(m-n)=a^m/a^n
->By using the above formula the problem can be solved easily
->(1/1+(a^n/a^m))+(1/1+(a^m/a^n))
->Take LCM of denominators of each terms individually
->(1/((a^m+a^n)/a^m))+(1/((a^n+a^m)/a^n))
->(a^m/(a^m+a^n))+(a^n/(a^n+a^m))
->By taking LCM, both numerator and denominator gets cancelled by 1.
->The answer is 1

Like? Yes (17) | No (4)  

Amit chaturvedi   9 years AGO

suppose a=1,m=2,n=3
put these values in equation
solving it
(1 / 1+1^(3-2)) + (1 / 1+1^(2-3))

1

Like? Yes (4) | No   

anjana sethi   9 years AGO

1/1+a^n-m +1/1+a^m-n = a^n +a^m/a^n+a^m = 1

Like? Yes (5) | No (1)  

J E Tejeswar Gowd   9 years AGO

1/(1+a^(m-n))+(a^(m-n))/(1+a^(m-n))=1

Like? Yes (5) | No (2)  

suraj kumar   9 years AGO

a^m+ a^n ÷a^m+a^n=1

Like? Yes (4) | No (1)  

ramya   9 years AGO

Take the first term:
1/1+a^(n-m) => 1/1+a^-(m-n)
=> a^(m-n) /1+a^(m-n) is the first term
Add it with second term you get :

1+a^(m-n) / 1+a^(m-n) => 1
So option 2

Like? Yes (3) | No   

Animesh Gupta   9 years AGO

for any value of n and m. the condition satisfies a^n+1/a^n+1.
which equals to 1.

Like? Yes (3) | No   

MURUGAVEL   9 years AGO

1/1+a^(n-m) = a^m/a^m+a^n
similarly
1/1+a^(m-n) = a^n/a^m+a^n
substituting in question we get ans 1,(i.e) option 2

Like? Yes (6) | No (3)  

praveendas   9 years AGO

the given expression can be rewritten like (a^m/a^m+a^n)+(a^n/a^m+a^n).
take lcm then numerator and denominator gets cancelled.that's it
option 2

Like? Yes (3) | No (1)  

kriti kankariya   9 years AGO

1/(1+((a^n)/(a^m))) = a^m/(a^m + a^n)
1/(1+((a^m)/(a^n))) = a^n/(a^m + a^n)
adding we will get 1

Like? Yes (4) | No (2)  

sudiptaa   9 years AGO

(1+a^m-n +1+a^n-m)/(1+a^m-n)(1+a^n-m))=2

Like? Yes (2) | No   

SUMIT PAREEK   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n)) =(1+a^(n-m)+1+a^(m-n))/((1+a^(n-m))*(1+a^(m-n)))
=(2+a^(m-n)+a^(n-m))/(2+a^(m-n)+a^(n-m))
=1

Like? Yes (4) | No (2)  

Rohit Agrawal   9 years AGO

affter addition the numerator and denominator pairs will cancel out

Like? Yes (2) | No   

shivam dubey   9 years AGO

2+ a^(m-n)+a^(n-m) / 2+a^(m-n)+a^(n-m) = 1 ans

Like? Yes (2) | No   

SANJIT PRASAD   9 years AGO

assume n-m=x;
1+a^x/1+a^x=1

Like? Yes (3) | No (1)  

Moumita Choudhury   9 years AGO

1/ (1+k) + 1 / (1+1/k) =1

Like? Yes (3) | No (1)  

Arvind kumar   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1 / 1+a^(n-m))+(1/1+(1/a^(n-m)))
=(1 / 1+a^(n-m))+(a^(n-m)/(1+a^(n-m)))
=(1 / 1+a^(n-m))/(1 / 1+a^(n-m))
=1

Like? Yes (2) | No   

himanshu shekhar   9 years AGO

(1/(1+a^n-m)+(1/(1+a^m-n)
=let n-m=x
ymthen (1/a^x)+(1/a^-x)
=1

Like? Yes (2) | No   

Deepali Tyagi   9 years AGO

This is of the form x / (x+1)+1/(x+1) where x=a^(m-n)
Option 2 is correct

Like? Yes (2) | No   

Ajay Kumar Gupta   9 years AGO

(1/(1+a^m/a^n))+(1/(1+a^n/a^m))
=a^m/(a^m+a^n)+a^n/(a^m+a^n)
=(a^m+a^n)/(a^m+a^n)
=1.

Like? Yes (2) | No   

Kaviya   9 years AGO

answer is 1
(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
by taking lcm :
1+a^(m-n)+1+a^(n-m)
=-------------------------------------
(1+a^(m-n))*(1+a^(n-m))

here a^(m-n) = a^-(n-m)= 1/ a^(n-m)
apply this in above equation
2+ a^(n-m)+ (1/a^(n-m))
=-----------------------------------------
(1+a^(n-m))*(1+ (1/a^(n-m)))

2a^(n-m)+1+ [ a^(n-m)*a^(n-m)]
=-----------------------------------------------
[ 1+a^(n-m)]^2

2a^(n-m)+1+ a^2(n-m)
=-----------------------------------------------
[ 1+a^(n-m)]^2

The numerator is of the form a^2 +b^2 +2ab = (a+b)^2
2a^(n-m)+1+ a^2(n-m) = [ 1+a^(n-m)]^2

[ 1+a^(n-m)]^2
=------------------------------------
[ 1+a^(n-m)]^2
=1

Like? Yes (2) | No   

bhaskar   9 years AGO

(1 / 1+a^(m-n))=a^(n-m)/1+a^(n-m)

Like? Yes (2) | No (1)  

SAGNIK DATTA   9 years AGO

case 1 : take n=2,m=1, a=2
case 2 : take n=3, m=1, a=2
both cases answer is 1

Like? Yes (2) | No (1)  

PRAGATHI G   9 years AGO

first term results as a^m/(a^m+a^n). second term results as a^n/(a^n+a^m). so addition of both terms results as 1

Like? Yes (3) | No (2)  

TAUSEEF AMEEN   9 years AGO

take lcm of (1+a^(n-m)) and (1+a^(m-n))..
will be same as numerator in addition..
hence 1.

Like? Yes (2) | No (1)  

RAVAL   9 years AGO

assume m=4,n=3,a=2

Like? Yes (2) | No (1)  

alpesh kurhade   9 years AGO

By solving simply we get ans: 1

Like? Yes (2) | No (1)  

Abhinav Akash   9 years AGO

Simple Fractional sum .

Like? Yes (2) | No (1)  

Danda Sindhura   9 years AGO

LET m,n be 1,2
by substituting in the given problem
we get numerator and denominator same ...
so ans is 1

Like? Yes (1) | No   

shikha   9 years AGO

1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=1+a^(m-n)+1+a^(n-m)/2+a^(m-n)+a^(n-m)
=1

Like? Yes (1) | No   

Sruthi Sreejan   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))

= (1 / 1+a^ -(m-n)) + (1 / 1+a^(m-n))
=(a^(m-n) /1+a^(m-n)) + (1 / 1+a^(m-n))
=(1+a^(m-n) / 1+a^(m-n))
=1

Like? Yes (1) | No   

shilpa singh   9 years AGO

1/1+a^n-m+1/1+a^m-n=1

Like? Yes (1) | No   

Gaurav Pratap Singh   9 years AGO

given question is (1/(1+a^n-m))+(1/(1+a^m-n))
we can write it as (1/(1+a^n-m))+(1/(1+ a^ (n-m)^ -1 ) )
=(1+a^n-m)/(1+a^n-m)
=1

Like? Yes (1) | No   

Arnab Maity   9 years AGO

= (1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
= (1 / 1+a^-b) + (1 / 1+a^b) [let m-n = b therefore -(m-n)= -b]
= (1 / 1+(1/a^b)) + (1 / 1+a^b)
= (a^b/1+a^b) + (1 / 1+a^b)
= (1 + a^b)/(1 + a^b)
= 1

Like? Yes (1) | No   

K.THARUN KUMAR   9 years AGO

(a^(m+n)/a^(m+n)+1)+(1/a^(m+n)+1)=1

Like? Yes (1) | No   

SK INTAJ MOHAMMAD   9 years AGO

a^m/(a^m+a^n)+a^n/(a^n+a^m)=1;

Like? Yes (1) | No   

Ranjana Gautam   9 years AGO

1/1+a^(n-m)+1/1+a^(m-n)
=a^(m-n)/1+a^(m-n)+1/1+a^(m-n)
=1

Like? Yes (1) | No   

sumitha saseendran   9 years AGO

a^n-m = a^n / a^m
similarly a^m-n = a^n / a^m
cross multiplying the two terms in the question and cancelling out common terms in numerator and denominator of the resulting fraction will finally give 1, i.e option 2

Like? Yes (1) | No   

meghna jain   9 years AGO

we wil write a^n/a^m and vice versa in other part and wil take the lcm . it will give 1 as ans

Like? Yes (1) | No   

k. asha   9 years AGO

a^m/(a^m+a^n)+a^n/(a^n+a^m)=1

Like? Yes (1) | No   

Deepak   9 years AGO

Let x = n-m,

1/ (1+a^x) + 1/(1+a^-x) = 1/(1+a^x) + a^x / (1+ a^x) = 1

Like? Yes (1) | No   

Praveenkumar Kalikeri   9 years AGO

Let m-n =p
then
1/( 1 +a^-p) + 1 / (1+a^p)
??
a^p /(1 +a^p) + 1/( 1 +a^p)
??
(1 +a^p)/(1+a^p)
?? =1
Hence option( 2 ). 1

Like? Yes (1) | No   

Rupkatha Banerjee   9 years AGO

given expression = (a^m/(a^m+a^n)) + (a^n/(a^m+a^n))
= (a^m+a^n)/(a^m+a^n)
=1

Like? Yes (1) | No   

Pradeep   9 years AGO

numerator and denominator will be cancelled

Like? Yes (1) | No   

Ankita Sharma   9 years AGO

m=0,n=0

Like? Yes (1) | No   

Ramesh Sriram   9 years AGO

by taking LCM and simplification

Like? Yes (1) | No   

UJJWAL KAMAR   9 years AGO

a^m/(a^m+a^n) + a^n/(a^m+a^n)

= (a^m+a^n)/(a^m+a^n)

=1

Like? Yes (1) | No   

Kanchan Patil   9 years AGO

(1/(1+(a^n/a^m)))+(1/(1+(a^m/a^n)))=
(a^m/(a^m+a^n))+(a^n/(a^n+a^m))=
(a^m+a^n)/(a^m+a^n)=1

Like? Yes (1) | No   

jyoti   9 years AGO

solving first of addends by reciprocating the "a" term, we get:
(1 / 1+(1/a^(m-n))

Simplifying it further we get:
(a^(m-n) / 1 + a^(m-n))

now considering the whole addition, that is:
(a^(m-n) / 1 + a^(m-n)) + (1 / 1+a^(m-n))

we get:
(a^(m-n) + 1 ) / ( 1 + a^(m-n))
which eventually results in 1.

So correct option is (2)

Like? Yes (1) | No   

sanchit soni   9 years AGO

1/1+a^n-m + 1/1+a^m-n =
1+a^m-n + 1+a^n-m/1+a^m-n + 1+a^n-m = 1 ans

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

[1/(1+a^(n-m))] + [1/(1+a^(m-n))]
As (n-m)=-(m-n)
So [1/(1+a^-(m-n))]+[1/(1+a^(m-n))]
If a^(m-n)=x ,then
[1/(1+x^-1)] +[1/(1+x)]
=1/(1+(1/x)) + 1/(1+x)
=x/(1+x) + 1/(1+x)
=(1+x)/(1+x)
=1
Option 2)

Like? Yes (1) | No   

MUNISH YADAV   9 years AGO

For n=m=0,1 values of expression is 1 for all a.

Like? Yes (1) | No   

Akash Kumar   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
adding both terms
=(1+a^(n-m)+1+a^(m-n))/((1+a^(n-m))*(1+a^(n-m)))
(solving numerator and denominator we get the following
=(2+a^(n-m)+a^(m-n))/(2+a^(n-m)+a^(m-n))
=1

Like? Yes (1) | No   

Dhrubo   9 years AGO

== [ 1 / {1 a^ n / a^ m} ] [ 1 / {1 a^ m / a^ n} ]
== [{a^ m / (a^m a ^ n)} {a^ n / (a^m a ^ n)} ]
== (a^ m a ^ n) / (a^ m a ^ n)

== 1 answer

Like? Yes (1) | No   

Bhavya Manasa   9 years AGO

(1/1+a^(n-m))+(1/1+a^(-(n-m))
(1/1+a^(n-m))+(1/(1+1/a^(n-m)))
(1/1+a^(n-m))+(a^(n-m)/1+a^(n-m))
(1+a^(n-m))/(1+a^(n-m))
1

Like? Yes (1) | No   

Arpan Ganguly   9 years AGO

1+a^(n-m) = 1+(a^n/a^m) = (a^m+a^n)/a^m and
1+a^(m-n) = 1+(a^m/a^n) = (a^m+a^n)/a^n
so
(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))=
(a^m/(a^m/a^n))+(a^n/(a^m/a^n))= ((a^m+a^n)/(a^m+a^n))=1

Like? Yes (1) | No   

shivam gaba   9 years AGO

1/1+an/am+1/1+am/an=1

Like? Yes (1) | No   

abhishek   9 years AGO

just simplify it...

firt take LCM
then you get denominator same as that of numerator
so equals 1

Like? Yes (1) | No   

pratik raj   9 years AGO

a^n-m will get cancelled if we take - common from second term

Like? Yes (1) | No   

Sriparna Boote   9 years AGO

Lets assume m=1 and n=2
Then
1/(1+a^-1)+1/(1+a)=1/(1+(1/a))+1/(1+a)=a/(1+a)+1(/(1+a)=(a+1)/(1+a)=1

Like? Yes (1) | No   

Debashish Mishra   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1 / 1+a^(n-m)) + (1 / 1+a^-1(n-m))
=(1 / 1+a^(n-m)) + (a^(n-m)/ 1+a^(n-m))
=(1+a^(n-m))/(1+a^(n-m))
=1

Like? Yes (1) | No   

avinash choori   9 years AGO

take a=4 m=2 and n=3
1/1+4^3-2 + 1/1+4^2-3=1

Like? Yes (1) | No   

prashant   9 years AGO

after solving the equation we get :
(2+a^(m-n)+a^(n-m))/(2+a^(m-n)+a^(n-m))=1

Like? Yes (1) | No   

Umesh sahu   9 years AGO

because they are cancelled all together when give any value assume in m and n

Like? Yes (1) | No   

YASWANTH REDDY   9 years AGO

if we simplify the given expression, we finally get same value in numerator and denominator and therefore the answer is 1

Like? Yes (1) | No   

adarsh   9 years AGO

on solving the above we get 1

Like? Yes (1) | No   

meghana   9 years AGO

a^(n-m) =p and a^(m-n) = 1/p

(1/(1+P))+(P/(P+1)) = (p+1)/(p+1) =1

Like? Yes (1) | No   

baby posamma   9 years AGO

(1/1+a^(n-m))+(a^(n-m)/(1+a^(n-m))=1

Like? Yes (1) | No   

sushil singh   9 years AGO

addition by taking lcm

Like? Yes (1) | No   

Akhila   9 years AGO

1/1+a*(n-m)+1/1+1/a*(n-m)

Like? Yes (1) | No   

keerthi choudhry   9 years AGO

1 / 1+a^(n-m) + 1 /1+a^(m-n) = 1/(1+a^n/a^m) + 1/(1+a^m/a^n)
= a^m/ a^m+a^n + a^n/ a^n+a^m
= a^n+a^m / a^n+a^m
=1(option 2)

Like? Yes (1) | No   

mithilesh kumar   9 years AGO

given (1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=>a^m/a^n+a^m +a^n/a^m+a^n
=>1

Like? Yes (1) | No   

payal   9 years AGO

on solving it looks like (2+a^m-n+a^n-m)/(2+a^m-n+a^n-m)=1

Like? Yes (1) | No   

Diksha tiwari   9 years AGO

1/(1+a^(n-m))+ 1/(1+a^(m-n)) =1/(1+a^(n-m))+a^(n-m)/(1+a^(n-m))
=(1+a^(n-m))/(1+a^(n-m))=1

Like? Yes (1) | No   

Maneesh Shringi   9 years AGO

(1/1+a^n-m) + (1/1+a^m-n)= 1/1+(a^n/a^m)+1/1+(a^m/a^n)
=a^m/(a^m+a^n) + a^n/(a^n+a^m)
=a^m+a^n/a^m+a^n
=1

Like? Yes (1) | No   

Anand Kumar Mehta   9 years AGO

After adding
Numerator becomes
(1+a^(m-n)+1+a^(n-m))
And denominator becomes same 1+a^(m-n)+a^(n-m)+a(n-m+m-n) which is equal to numarator ......so ans.is 1 option 2 is correct

Like? Yes (1) | No   

Subhendu Pradhan   9 years AGO

(1/1+a^n/a^m)+(1/1+a^m/a^n)
=a^m/(a^m+a^n)+a^n/(a^m+a^n)
=a^m+a^n/a^m+a^n
=1

Like? Yes (1) | No   

s.h.h.praveen   9 years AGO

just by simplification and a formula (a^n-m=a^n/a^m)

Like? Yes (1) | No   

kondapuram tirumalreddy   9 years AGO

by solving we get
2+a^n-m+a^m-n/2+a^m-n+a^n-m
=1

Like? Yes (1) | No   

Aman Gaurav   9 years AGO

If base are same divide power should be subtracted
so,
1/1+(a^n/a^m) + 1/1+(a^m/a^n)
so,
a^m/a^m+a^n + a^n/a^m+a^n
so,
a^m+a^n/a^m+a^n
so
1

Like? Yes (1) | No   

Aman   9 years AGO

If base are same divide power should be subtracted
so,
1/1+(a^n/a^m) + 1/1+(a^m/a^n)
so,
a^m/a^m+a^n + a^n/a^m+a^n
so,
a^m+a^n/a^m+a^n
so
1

Like? Yes (1) | No   

lakshmi priya   9 years AGO

on taking lcm & canceling

Like? Yes (1) | No   

S.RAJASHREE   9 years AGO

PUTa=3,n=6,m=5,solving we get 1

Like? Yes (1) | No   

Vikas Jha   9 years AGO

Solving it gets to 1

Like? Yes (1) | No   

akhil   9 years AGO

change a^(n-m)as 11/a^(m-n) and take l.c.m

Like? Yes (1) | No   

Pesari Udayasree   9 years AGO

let n-m=b; then m-n=-b
by substituting this values we get the equation as
(1/1+a^b)+(1/1+a^-b)
since a^-b=1/a^b
then we get (1+a^b)/(1+a^b)=1

Like? Yes (1) | No   

Nick Jain   9 years AGO

(2+a^(n-m)+a^(m-n))/(2+a^(n-m)+a^(m-n))=1

Like? Yes (1) | No   

Mayank   9 years AGO

(1/1+a^2)+(1/1+a^-2)
=(1/1+a^2)/(1/1+a^2)
=1

Like? Yes (1) | No   

mounika   9 years AGO

(1/1+a^2)+(1/1+a^-2)
=(1/1+a^2)/(1/1+a^2)
=1

Like? Yes (1) | No   

jagdish jakhar   9 years AGO

[1+a^(m-n)+1+a^(n-m)]/2+a^(m-n)+a^(n-m)=1

Like? Yes (1) | No   

Anand Kumar Sharma   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1/1+a^(n-m))+(a^(n-m)/1+a^(n-m))
=(1+a^(n-m))/(1+a^(n-m))
=1

Like? Yes (1) | No   

krishna prasad   9 years AGO

after simplification numerator and denominator are equal

Like? Yes (1) | No   

Anubrindha   9 years AGO

Use simple cross multiplication we can get 1 as final answer

Like? Yes (1) | No   

Surabhi Yadav   9 years AGO

by solving it (1/1+a^n-m)+(1/1+a^m-n)
=(1/1+(a^n/a^m))+(1/1+(a^m/a^n)
=(a^m/(a^m+a^n))+(a^n/(a^m+a^n))
=1

Like? Yes (1) | No   

K.Anitha   9 years AGO

When simplified equation becomes as
(2+a^(m-n)+a^(n-m))/(1+a^(m-n)+a^(n-m)+a^0)=1

Like? Yes (1) | No   

prashant trivedi   9 years AGO

simple just solve

Like? Yes (1) | No   

surya   9 years AGO

both numerator and denominator are same after solving

Like? Yes (1) | No   

rahul shaw   9 years AGO

by solving the equations we get answer is 1
Just we have to take -ve sign common frpm 2nd term and simplify.

Like? Yes (1) | No   

Kunal nemu   9 years AGO

1 / 1+(a^n/a^m) + 1 / 1+(a^m/a^n)
=1 / ((a^n+a^m)/a^m) + 1 / ((a^n+a^m)/a^n)
=a^m/(a^n+a^m) + a^n/(a^n+a^m)
=(a^n+a^m) / (a^n+a^m)
=1

Like? Yes (1) | No   

john pranay   9 years AGO

a^m+a^n/a^m+a^n=1

Like? Yes (1) | No   

vimal tiwari   9 years AGO

1/1+a^m/a^n => a^n / a^n +a^m
and next term is solved as same., it becomes
a^m/a^n+a^m

on summing both we get =1

Like? Yes (1) | No   

Arvind   9 years AGO

(a^(m)/(a^(m)+a^(n)))+(a^(n)/(a^(m)+a^(n)))

Like? Yes (1) | No   

Mugesh sivagnanam Krishnapillai   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=[2+a^(m-n)+a^(n-m)] / [1+a^(n-m))*(1+a^(m-n)]
=[2+a^(m-n)+a^(n-m)]/[2+a^(m-n)+a^(n-m)]
=1.

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velmala gangadhar   9 years AGO

(1/1+a^(n-m))+1/(1+1/a^n-m)..
(1/1+a^(n-m))+a^n-m/(1+a^n-m)..
it will become 1......

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MADHUBABU   9 years AGO

=>1/(1+(a^n/a^m))+1/(1+(a^m/a^n))
=>a^m/(a^m+a^n) +a^n/(a^n+a^m)
=>(a^m + a^n)/(a^m + a^n)
=> 1.

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jayaram   9 years AGO

a^m/(a^m+a^n)+a^n/(a^m+a^n)=1

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Priya Srimani   9 years AGO

= 1/{1+ (a^n+a^m)} + 1/1+(a^m+a^n)}
=1/{(a^m+a^n)/ a^m} + 1/{(a^n+a^m)/a^n}
= {a^m/(a^m+a^n)} + {a^n/(a^n+a^m)}
= (a^m+a^n)/(a^m+a^n)
= 1

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Sharath   9 years AGO

1/1+a^(n-m) + 1/1+a^-(n-m)
if we solve this we get
1/1+a^(n-m) + a^(n-m)/1+a^(n-m) = 1
so, option 2.

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saqib khan   9 years AGO

let n-m=x
then, m-n=-x
take LCM and on solving numerator and denominator will get cancelled and we will get 1.

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ramakrishna   9 years AGO

(a^m/(a^m+a^n))+(a^n/a^m+a^n))=
(a^m+a^n)/(a^m+a^n)=1

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CHANDAN KUMAR   9 years AGO

(a^m/a^m+a^n) +(a^n /a^n+a^m) =1

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Parthasarathi.S   9 years AGO

m-n=x
[1/(1+a^-x)]+ [1/(1+a^x)]
=[(a^x)/(1+a^x)] + [1/(1+a^x)]
=(1+a^x)/(1+a^x)
=1

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Saurabh Agrawal   9 years AGO

(1 / 1+a^(m-n)) = (n-m) / 1+a^(n-m)
(1 / 1+a^(n-m))+((n-m) / 1+a^(n-m)) = 1

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MANUPOTI ANAND   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))=(1/(1+(1/a^m-n)))+ (1 / 1+a^(m-n))
=((a^m-n)/ (1 / 1+a^(m-n)))+ (1 / 1+a^(m-n))
= (1 / 1+a^(m-n))/ (1 / 1+a^(m-n))
=1

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RUPANJAN ROY   9 years AGO

1/(1+(a^n/a^m))+1/(1+(a^m/a^n))
this sums up to 1

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Gowsalaya G   9 years AGO

(1/(1+a^(n-m))+(1/1+a^(m-n))
=(1+a^(m-n)+1+a^(n-m))/((1+a^(n-m))(1+a^(m-n)))
=(2+a^(m-n)+a^(n-m))/(2+a^(m-n)+a^(n-m))
=1

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Animesh   9 years AGO

let m-n=x ,then 1/a^x + 1/a^-x =1

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Nandkishor jaiswal   9 years AGO

(1/1+a^(n-m))+(1/1+a^-(n-m))
=(1/1+a^(n-m)+(a^(n-m))/1+a^(n-m))
=1+a^(n-m)/1+a^(n-m)
=1

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sukriti singh   9 years AGO

1/(1+a^m-n)+1/(1+a^n-m)
=(1+a^m-n+1+a^m-n)/(1+a^m-n)*(1+a^-(m-n))
=(2+a^m-n+a^-(m-n))/(1+a^m-n)*(1+a^-(m-n))
=(2*a^m-n+1+a^2m-n)/a^m-n*(1+a^m-n)*(1+a^-(m-n))
=(a^2m-n+1)^2/(1+a^m-n)*(1+a^m-n)
=we will get 1 solving this..
so ans is option 2.

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rajat gupta   9 years AGO

a^m+a^n/a^m+a^n=1

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shafi   9 years AGO

no need of much explanation for this

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Ishita Chakraborty   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1 / 1+a^(n-m)) + (1 / 1+a^ -(n-m))
=(1 / 1+a^(n-m)) + (1 / 1+(1/a^(n-m)))
=(1 / 1+a^(n-m))+ (a^(n-m)/1+a^(n-m))
=(1+a^(n-m))/(1+a^(n-m))

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Subhasri Pal   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1 / 1+a^(n-m)) + (1 / 1+(1/a^(n-m)))
=a^(n-m)+1/a^(n-m)+1
=1

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kumar ajay   9 years AGO

1/(1+(a^n/a^m))+1/(1+(a^m/a^n))
a^m/(a^m+a^n)+a^n/(a^m+a^n)=1

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Shubham Raj   9 years AGO

the correct ans is 1
option 2)1

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Deepanshu Chaturvedi   9 years AGO

let n-m=y
so, from equation
(1/(1+a^y))+(1/(1+a^-y))
=> (1+a^y)/(1+a^y)=1

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Mayank Aditya   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
=(1 / 1+a^(n-m))+ (a^(n-m) / 1+a^(n-m))
=1

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Deepak Agarwal   9 years AGO

let n=1, m=2

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Kethineni Vinodh   9 years AGO

on expanding num/dem=1

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SAI TEJA NALLA   9 years AGO

(1 / 1+a^(n-m)) + (1 / 1+a^(m-n))
(1 / 1+a^(n-m)) + (a^(n-m) / 1+a^(n-m))
1+a^(n-m)/1+a^(n-m)=>1

Like? Yes (1) | No (1)  

Nizamuddin S   9 years AGO

(a^(m+n)/a^(m+n)+1) + (1/a^(m+n)+1) =1

Like? Yes (1) | No (1)  

nirmal das   9 years AGO

1+a^(n-m)=(a^m+a^n)/a^m
in this way another part is solved....then if we add their reciprocal then we get 1..

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pavani mongam   9 years AGO

(1+a^(m-n)+1+a^(n-m))/(1+a^(n-m))(1+a^(m-n))
=2+a^(m-n)+a^(n-m)/(2+a^(m-n)+a^(m-n))
=1

Like? Yes (1) | No (1)  

preethy vijayakumar   9 years AGO

1/[1+(1/a^m-n)]+1/[1+a^m-n]
=(a^m-n)+1/(a^m-n)+1
=1

Like? Yes (1) | No (1)