Prove that: \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=\sec A\text{cosec}A+1\]

Hint: We will begin with the left hand side of the equation (1) and then we will convert all the cot terms in tan and when we will proceed further we will get \[{{1}^{3}}-{{\tan }^{3}}A\] which we will expand using the formula \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and finally we will get the answer which will be equal to the right hand side of the expression. We will use the formula \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] which will help us in proving the given expression.

Complete step-by-step answer:

It is mentioned in the question that \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=\sec A\text{cosec}A+1.......(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow \dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}.........(2)\]

Now simplifying by converting all the cot terms in equation (2) and hence we get,

\[\Rightarrow \dfrac{\tan A}{1-\dfrac{1}{\tan A}}+\dfrac{\dfrac{1}{\tan A}}{1-\tan A}.........(3)\]

Now we will take the LCM in the first term of the equation (3) and simplifying second term we will get,

\[\Rightarrow \dfrac{{{\tan }^{2}}A}{\tan A-1}+\dfrac{1}{\tan A(1-\tan A)}.........(4)\]

Now rearranging equation (4) we get,

\[\Rightarrow \dfrac{1}{\tan A(1-\tan A)}-\dfrac{{{\tan }^{2}}A}{1-\tan A}.........(5)\]

Now again taking the LCM in equation (5) we get,

\[\Rightarrow \dfrac{1-{{\tan }^{3}}A}{\tan A(1-\tan A)}.........(6)\]

Now we know that \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\]. So applying this formula in equation (6) we get,

\[\Rightarrow \dfrac{(1-\tan A)(1+{{\tan }^{2}}A+\tan A)}{\tan A(1-\tan A)}.........(7)\]

Now cancelling similar terms in equation (7) we get,

\[\Rightarrow \dfrac{1+{{\tan }^{2}}A+\tan A}{\tan A}.........(8)\]

Now again substituting \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\] in equation (8) we get,

\[\Rightarrow \dfrac{{{\sec }^{2}}A+\tan A}{\tan A}.........(9)\]

Now dividing the numerator terms separately and transforming the terms in sin, cos in equation (9) we get,

\[\begin{align}

  & \Rightarrow \dfrac{{{\sec }^{2}}A}{\tan A}+\dfrac{\tan A}{\tan A} \\

 & \Rightarrow \dfrac{\cos A}{{{\cos }^{2}}A\sin A}+1......(10) \\

\end{align}\]

Now cancelling the similar terms in equation (10) and getting the terms in cosec and sec we get,

\[\begin{align}

  & \Rightarrow \dfrac{1}{\cos A\sin A}+1 \\

 & \Rightarrow \sec A\text{cosec}A+1.......(11) \\

\end{align}\]

Hence from equation (11) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (9) but here the key is to convert all the terms in sin and cos. Also we will use \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] a few times to get the left hand side equal to the right hand side of the equation (1).