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

The following expression can be written as:
tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}

Explanation

Solution

Hint: In order to solve this problem you need to use the formulas tan⁡A=sin⁡Acos⁡A   and cot⁡A=cos⁡Asin⁡A\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 tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}.
Putting tan⁡A=sin⁡Acos⁡A   and cot⁡A=cos⁡Asin⁡A\tan A = \dfrac{{\sin A}}{{\cos A}}\,\,\,{\text{and }}\cot A = \dfrac{{\cos A}}{{\sin A}} the above equation can be written as:
⇒sin⁡Acos⁡A1−cos⁡Asin⁡A+cos⁡Asin⁡A1−sin⁡Acos⁡A\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,
⇒sin⁡Acos⁡Asin⁡A−cos⁡Asin⁡A+cos⁡Asin⁡Acos⁡A−sin⁡Acos⁡A ⇒sin⁡2Acos⁡A(sin⁡A−cos⁡A)+cos⁡2Asin⁡A(cos⁡A−sin⁡A) ⇒sin⁡3A−cos⁡3Asin⁡Acos⁡A(sin⁡A−cos⁡A)  \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 a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})
⇒(sin⁡A−cos⁡A)(sin⁡2A+cos⁡2A+sin⁡Acos⁡A)sin⁡Acos⁡A(sin⁡A−cos⁡A) ⇒1+sin⁡Acos⁡Asin⁡Acos⁡A      (As sin2x+cos⁡2x=1)  \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 1sin⁡x=cos⁡ecx    &    1cos⁡x=sec⁡x\dfrac{1}{{\sin x}} = \cos ecx\,\,\,\,\& \,\,\,\,\dfrac{1}{{\cos x}} = \sec x.
Further simplifying the equation 1+sin⁡Acos⁡Asin⁡Acos⁡A \dfrac{{1 + \sin A\cos A}}{{\sin A\cos A}}\, we get,
⇒1+sin⁡Acos⁡Asin⁡Acos⁡A =1sin⁡Acos⁡A+sin⁡Acos⁡Asin⁡Acos⁡A ⇒sec⁡Acos⁡ecA+1  \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 tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} can also be written as sec⁡Acos⁡ecA+1\sec A\cos ecA + 1.

Note: In this question you just have to use the formulas tan⁡A=sin⁡Acos⁡A   and cot⁡A=cos⁡Asin⁡A\tan A = \dfrac{{\sin A}}{{\cos A}}\,\,\,{\text{and }}\cot A = \dfrac{{\cos A}}{{\sin A}} and also sin2x+cos⁡2x=1{\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.