Question

How do you prove $\dfrac{\cot x}{1-\tan x}+\dfrac{\tan x}{1-\cot x}=1+\tan x+\cot x$?

Answer 1

Hint: To solve this question we will use the trigonometric property that $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$. We will substitute the value in the LHS of the given expression and then simplify the expression to get the value equal to the RHS.

Complete step by step answer:
We have been given an expression $\dfrac{\cot x}{1-\tan x}+\dfrac{\tan x}{1-\cot x}=1+\tan x+\cot x$
We have to prove that LHS=RHS.
Let us first consider the LHS of the given expression we have $\dfrac{\cot x}{1-\tan x}+\dfrac{\tan x}{1-\cot x}$
Now, let us express the value of $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$.
Now, we know that $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$
Substituting the values in the given expression we get
$\Rightarrow \dfrac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}+\dfrac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}$
Now, simplifying the obtained equation we get
$\Rightarrow \dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x-\sin x}{\cos x}}+\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{\sin x-\cos x}{\sin x}}$
Now, rearranging the terms we get
\[\Rightarrow \dfrac{{{\cos }^{2}}x}{\sin x\left( \cos x-\sin x \right)}+\dfrac{{{\sin }^{2}}x}{\cos x\left( \sin x-\cos x \right)}\]
Now, we can rewrite the above equation as
\[\Rightarrow \dfrac{{{\cos }^{2}}x}{-\sin x\left( \sin x-\cos x \right)}+\dfrac{{{\sin }^{2}}x}{\cos x\left( \sin x-\cos x \right)}\]
Now, taking the common term out we get
\[\Rightarrow \dfrac{1}{\left( \sin x-\cos x \right)}\left[ \dfrac{{{\cos }^{2}}x}{-\sin x}+\dfrac{{{\sin }^{2}}x}{\cos x} \right]\]
Now, simplifying the obtained equation further we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\left( \sin x-\cos x \right)}\left[ \dfrac{-{{\cos }^{3}}x+{{\sin }^{3}}x}{\sin x\cos x} \right] \\
 & \Rightarrow \dfrac{1}{\left( \sin x-\cos x \right)}\left[ \dfrac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x\cos x} \right] \\
\end{align}\]
Now, we know that \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]
Now, by applying the formula in the above obtained equation we get
\[\Rightarrow \dfrac{1}{\left( \sin x-\cos x \right)}\left[ \dfrac{\left( \sin x-\cos x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x+\sin x\cos x \right)}{\sin x\cos x} \right]\]
Now, cancel out the common terms we get
\[\Rightarrow \left[ \dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x+\sin x\cos x \right)}{\sin x\cos x} \right]\]
Now, we can rewrite the above obtained equation as
\[\Rightarrow \dfrac{{{\sin }^{2}}x}{\sin x\cos x}+\dfrac{{{\cos }^{2}}x}{\sin x\cos x}+\dfrac{\sin x\cos x}{\sin x\cos x}\]
Now, cancelling the common terms we get
\[\Rightarrow \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x}+1\]
Now we know that $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$
Substituting the values in the above obtained equation we get
\[\Rightarrow \tan x+\cos x+1\] which is equal to the RHS.
Hence we get $\dfrac{\cot x}{1-\tan x}+\dfrac{\tan x}{1-\cot x}=1+\tan x+\cot x$
Hence proved

Note: Alternatively one can consider the RHS of the given expression and simplify it to obtain a value equal to the LHS. But sometimes it is difficult to start with the RHS and arrive at the LHS value. Another way is to solve both LHS and RHS at some point and get the equal values.