Question

Proving Identities

Verify the identity$\frac{1-\cos x}{\sin x}+\frac{\sin x}{1-\cos x}=2\csc x$

Short answer

The expression$\frac{1-\cos x}{\sin x}+\frac{\sin x}{1-\cos x}=2\csc x$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{1-\cos x}{\sin x}+\frac{\sin {\displaystyle }x}{1-\cos x}=2\csc x$

Step 2. Concept used.

To begin, divide the phrase into two halves, such as LHS and RHS. Separately simplify both portions. When the results of both sections are the same, the expression is called an identity.

Step 3. Calculation.

Now, simplify LHS of $\frac{1-\cos x}{\sin x}+\frac{\sin x}{1-\cos x}=2\csc x$:

For this, assume LHS as y

$\frac{(\csc x-\cot x)}{(\sec x-1)}=y$

Now take reciprocal on both sides and multiply by $\frac{(\csc x+\cot x)}{(\csc x+\cot x)}$,

$\begin{array}{l}(\sec x-1)(\mathrm{cosec}x+\cot x)=\frac{1}{y}\\ (a+b)(a-b)=a^2-b^2\\ \left({\csc }^{2}x-{\cot }^{2}x\right)=1\end{array}$

Now simplify further,

$\begin{array}{l}\Rightarrow \cot x\sec x-\mathrm{cosec}x-\cot x+\sec x\cdot \mathrm{cosec}x\\ \Rightarrow \left(\frac{\cos x}{\sin x}\times \frac{1}{\cos x}-\frac{1}{\sin x}-\frac{\cos x}{\sin x}+\frac{1}{\cos x}\times \frac{1}{\sin x}\right)=\frac{1}{y}\\ \Rightarrow \frac{1+\cos x}{\sin x\cos x}-\frac{1}{\sin x}-\frac{\cos x}{\sin x}=\frac{1}{y}\\ \Rightarrow \frac{1+\cos x-\cos x-{\cos }^{2}x}{\sin x\cos x}=\frac{1}{y}\\ \Rightarrow \frac{1-{\cos }^{2}x}{\sin x\cos x}=\frac{1}{y}\\ \because {\sin }^{2}x+{\cos }^{2}x=1\\ \Rightarrow \frac{{\sin }^{2}x}{\sin x\cos x}=\frac{1}{y}\\ \Rightarrow \tan x=\frac{1}{y}\\ \Rightarrow \cot x=y\end{array}$

RHS of the equation is$\cot x$

Both RHS and LHS are equal. Hence it is an identity