Question

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

Answer 1

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.
* \[\csc A = \dfrac{1}{{\sin A}};\cot A = \dfrac{{\cos A}}{{\sin A}}\]

Complete step-by-step answer:
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.