Question

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

Solution

## $$\begin{array}{l} \dfrac { { \tan A } }{ { 1-\cot A } } +\dfrac { { \cot A } }{ { 1-\tan A } } \\ \Rightarrow \dfrac { { \dfrac { { \sin A } }{ { \cos A } } } }{ { 1-\dfrac { { \cos A } }{ { \sin A } } } } +\dfrac { { \dfrac { { \cos A } }{ { \sin A } } } }{ { 1-\dfrac { { \sin A } }{ { \cos A } } } } \\ \Rightarrow \dfrac { { { { \sin }^{ 2 } }A } }{ { \cos A\left( { \sin A-\cos A } \right) } } +\dfrac { { { { \cos }^{ 2 } }A } }{ { \sin A\left( { \sin A-\cos A } \right) } } \\ \Rightarrow \dfrac { { { { \sin }^{ 3 } }A-{ { \cos }^{ 3 } }A } }{ { \sin A\cos A\left( { \sin A-\cos A } \right) } } \\ \Rightarrow \dfrac { { \left( { 1+\sin A\cos A } \right) } }{ { \sin A\cos A } } \\ \Rightarrow \dfrac { 1 }{ { \sin A.\cos A } } +\dfrac { { \sin A\cos A } }{ { \sin A\cos A } } \\ \Rightarrow \sec A\cos ecA+1 \end{array}$$Mathematics

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