5 options and none works

Question

anonymous · Answer

$$(a+b)^{-1}(a ^{-1}+b ^{-1})^{-1} $$ is equal to
$$(a ^{2}+b ^{2})$$
$$(a+b)^{-1}$$
$$(ab)^{-1}$$
$$a ^{2}+b ^{2}$$
$$ab$$

anonymous · Answer

i got 
$$ab/(a ^{2}+b ^{2})$$

anonymous · Answer

help

anonymous · Answer

oh first 1 option $$(a+b)^{2}$$

asnaseer · Answer

can you please show how you got to your answer of $ab/(a ^{2}+b ^{2})$?

anonymous · Answer

$$[1/(a+b)][1/a+1/b]^{-1}$$
$$[1/(a+b)][ab/(a+b)]$$
$$ab/(a+b)(a+b)$$
$$ab/(a+b)^{2}$$
i cant see any of the options

asnaseer · Answer

your answer of $ab/(a+b)^{2}$ is correct.
although the original answer you wrote above was wrong: $ab/(a ^{2}+b ^{2})$

anonymous · Answer

sorry but how does this change the answer

asnaseer · Answer

$$(a^2+b^2)\ne(a+b)^2$$

anonymous · Answer

how does $$ab/(a+b)^{2}$$ match any of the options

asnaseer · Answer

it doesn't, so either the question is stated incorrectly or the answer options are all wrong.
the work you have done to solve the question above is definitely correct.

asnaseer · Answer

is this from a book or is it online?

anonymous · Answer

its a book on test preparation 'mathsdigest'

asnaseer · Answer

I would say that the book has a misprint then. if you can, you should contact the author or publishers to let them know.

anonymous · Answer

i think on the question,the 2nd brackets part should not have the -1 so that option (5) works

asnaseer · Answer

possibly...

anonymous · Answer

thanks 4 your time asnaseer

asnaseer · Answer

yw :)