Question

The following expression can be written as:
$\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$

Answer 1

Hint: In order to solve this problem you need to use the formulas $\tan A = \dfrac{{\sin A}}{{\cos A}}\,\,\,{\text{and }}\cot A = \dfrac{{\cos A}}{{\sin A}}$. Using these and simplifying we get the simplified equation.

Complete step-by-step answer:
The given equation is $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$.
Putting $\tan A = \dfrac{{\sin A}}{{\cos A}}\,\,\,{\text{and }}\cot A = \dfrac{{\cos A}}{{\sin A}}$ the above equation can be written as:
$ \Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{1 - \dfrac{{\cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 - \dfrac{{\sin A}}{{\cos A}}}}$
On solving it further we get,
$
   \Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}} \\
   \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A(\sin A - \cos A)}} + \dfrac{{{{\cos }^2}A}}{{\sin A(\cos A - \sin A)}} \\
   \Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\sin A\cos A(\sin A - \cos A)}} \\
$
We know that ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
$
   \Rightarrow \dfrac{{(\sin A - \cos A)({{\sin }^2}A + {{\cos }^2}A + \sin A\cos A)}}{{\sin A\cos A(\sin A - \cos A)}} \\
   \Rightarrow \dfrac{{1 + \sin A\cos A}}{{\sin A\cos A}}\,\,\,\,\,\,({\text{As si}}{{\text{n}}^2}x + {\cos ^2}x = 1) \\
$
And we also know that $\dfrac{1}{{\sin x}} = \cos ecx\,\,\,\,\& \,\,\,\,\dfrac{1}{{\cos x}} = \sec x$.
Further simplifying the equation $\dfrac{{1 + \sin A\cos A}}{{\sin A\cos A}}\,$ we get,
$
   \Rightarrow \dfrac{{1 + \sin A\cos A}}{{\sin A\cos A}}\, = \dfrac{1}{{\sin A\cos A}} + \dfrac{{\sin A\cos A}}{{\sin A\cos A}} \\
   \Rightarrow \sec A\cos ecA + 1 \\
$
Hence, the term $\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}$ can also be written as $\sec A\cos ecA + 1$.

Note: In this question you just have to use the formulas $\tan A = \dfrac{{\sin A}}{{\cos A}}\,\,\,{\text{and }}\cot A = \dfrac{{\cos A}}{{\sin A}}$ and also ${\text{si}}{{\text{n}}^2}x + {\cos ^2}x = 1$. Using these and simplifying we will get the simplified term as an answer to this question.