Question

Prove that$\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = 1.$

Hint: To prove this, we simplify both the fractions separately from $\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}}$ by using laws of indices given below. Since, we have the same base i.e. x, so we can apply the following laws.
Laws of indices: $(i)$ ${a^m} \cdot {a^n} = {a^{m + n}}$
$(ii)$ $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
$(iii)$ ${a^{ - n}} = \dfrac{1}{{{a^n}}}$

Simplify $\dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}}$ by using laws of indices.
Use the law of indices ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ and rewrite the terms.
$\therefore \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{1 + \dfrac{{{x^a}}}{{{x^b}}}}} + \dfrac{1}{{1 + \dfrac{{{x^b}}}{{{x^a}}}}}$
Find the least common denominator and take LCM in the denominator of both fractions.
$\Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{1}{{\dfrac{{{x^b} + {x^a}}}{{{x^b}}}}} + \dfrac{1}{{\dfrac{{{x^a} + {x^b}}}{{{x^a}}}}}$
Remember, that dividing by fraction is the same as flipping the fraction in the denominator and then multiplying it with the numerator.
So flip the fraction in the denominator and multiply it to the numerator.
$\Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b}}}{{{x^a} + {x^b}}} + \dfrac{{{x^a}}}{{{x^a} + {x^b}}}$
Take LCM of both the fractions.
$\Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = \dfrac{{{x^b} + {x^a}}}{{{x^a} + {x^b}}}$
Cancel the same terms from numerator and denominator.
$\Rightarrow \dfrac{1}{{1 + {x^{a - b}}}} + \dfrac{1}{{1 + {x^{b - a}}}} = 1$
Hence proved.

Additional Information: * When the base is the same then only we can apply the law of indices, be it multiplication of numbers with the same base or division of numbers with the same base.
* Always keep in mind that any number with power 0 will be equal to 1.
$\Rightarrow {a^0} = 1$
* If powers are in fraction form, then we can write them as
$\Rightarrow {a^{\dfrac{m}{n}}} = {({a^m})^{\dfrac{1}{n}}}$ which means we take nth root of the term inside the bracket, so we can write
$\Rightarrow {a^{\dfrac{m}{n}}} = \sqrt[n]{{{a^m}}}$
* Any number with the power m means that the number is multiplied to itself m number of times.
So, we can write ${x^m} = \underbrace {x \times x \times x........ \times x}_m$

Note: Students make mistake of assuming the values ${x^{b - a}},{x^{a - b}}$as some variables and try solving the fraction which is wrong way to proceed with the question. So, keep in mind we have to use formulas which make solutions easy and not more complex.