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

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, \[\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}.\]

Many students make the mistake of solving the question without converting into sin and cos, which makes our solution complex.