Question

# Prove that: $\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}.$

Hint: To prove this, we simplify the both sides separately by writing all trigonometric ratios in terms of $\sin A$ and $\cos A$, then rationalize the term in the fraction which has complex denominator and solve to simplest form. We show that both sides give the same values.
* $\csc A = \dfrac{1}{{\sin A}};\cot A = \dfrac{{\cos A}}{{\sin A}}$

First we solve the LHS of the equation.
Write all trigonometric ratios in $\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}}$ in terms of $\sin A$ and $\cos A$.
$\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\dfrac{1}{{\sin A}} - \dfrac{{\cos A}}{{\sin A}}}} - \dfrac{1}{{\sin A}}$
Taking LCM in the denominator of the first fraction.
$= \dfrac{1}{{\dfrac{{1 - \cos A}}{{\sin A}}}} - \dfrac{1}{{\sin A}}$
$= \dfrac{{\sin A}}{{1 - \cos A}} - \dfrac{1}{{\sin A}}$
Now we rationalize the first fraction by multiplying both numerator and denominator by $(1 + \cos \theta )$.
$= \dfrac{{\sin A}}{{1 - \cos A}} \times \dfrac{{1 + \cos A}}{{1 + \cos A}} - \dfrac{1}{{\sin A}}$

$= \dfrac{{\sin A(1 + \cos A)}}{{(1 + \cos A)(1 - \cos A)}} - \dfrac{1}{{\sin A}}$
Now we know from the property that $(a + b)(a - b) = {a^2} - {b^2}$
Using the formula solve denominator of first fraction where $a = 1,b = \cos A$
$= \dfrac{{\sin A(1 + \cos A)}}{{(1 - {{\cos }^2}A)}} - \dfrac{1}{{\sin A}}$
Now from the property ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we can write $1 - {\cos ^2}\theta = {\sin ^2}\theta$
So, we substitute the value of $1 - {\cos ^2}A = {\sin ^2}A$ in the denominator of the first fraction.
$= \dfrac{{\sin A(1 + \cos A)}}{{{{\sin }^2}A}} - \dfrac{1}{{\sin A}}$

Cancel out the same factors from numerator and denominator.

$= \dfrac{{(1 + \cos A)}}{{\sin A}} - \dfrac{1}{{\sin A}}$

Taking LCM of both the fractions.

$= \dfrac{{1 + \cos A - 1}}{{\sin A}} \\ = \dfrac{{\cos A}}{{\sin A}} \\$

$= \cot A$ {since $\cot A = \dfrac{{\cos A}}{{\sin A}}$ }

$\Rightarrow \dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \cot A$ $...(1)$
Now we solve the RHS of the equation.
Write all trigonometric ratios in $\dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}$ in terms of $\sin A$ and $\cos A$.
$\dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}}}$
Take LCM in the denominator of the second fraction.
$= \dfrac{1}{{\sin A}} - \dfrac{1}{{\dfrac{{1 + \cos A}}{{\sin A}}}}$
$= \dfrac{1}{{\sin A}} - \dfrac{{\sin A}}{{1 + \cos A}}$
Now we rationalize the second fraction by multiplying both numerator and denominator by $(1 - \cos A)$
$= \dfrac{1}{{\sin A}} - \dfrac{{\sin A}}{{1 + \cos A}} \times \dfrac{{1 - \cos A}}{{1 - \cos A}}$

$= \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{(1 - \cos A)(1 + \cos A)}}$
Now we know from the property that $(a + b)(a - b) = {a^2} - {b^2}$
Using the formula solve denominator of second fraction where $a = 1,b = \cos A$
$= \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{1 - {{\cos }^2}A}}$
Now from the property ${\sin ^2}\theta + {\cos ^2}\theta = 1$ we can write $1 - {\cos ^2}\theta = {\sin ^2}\theta$
So, we substitute the value of $1 - {\cos ^2}A = {\sin ^2}A$ in the denominator of the second fraction.
$= \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{{{\sin }^2}A}}$
Cancel out factors from the numerator and denominator of the second fraction.
$= \dfrac{1}{{\sin A}} - \dfrac{{(1 - \cos A)}}{{\sin A}}$
Taking LCM of both fractions.
$= \dfrac{{1 - 1 + \cos A}}{{\sin A}} \\ = \dfrac{{\cos A}}{{\sin A}} \\$
$= \cot A$ {since $\cot A = \dfrac{{\cos A}}{{\sin A}}$ }
$\Rightarrow \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}} = \cot A$ $...(2)$
Now we check if LHS is equal to RHS
Equating values of $\cot A$ from equation $(1)$ and $(2)$ we get
$\cot A = \cot A$
LHS = RHS
Hence, $\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}.$

Note: Students make mistakes of rationalizing with wrong factors, keep in mind we multiply with such a term that makes our denominator easy.
Many students make the mistake of solving the question without converting into sin and cos, which makes our solution complex.