Question

Proving Identities

Verify the identity$\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin {\displaystyle }\alpha }{1+\cos \alpha }$

Short answer

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

Step-by-step solution

Step 1. Given information.

An expression$\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin {\displaystyle }\alpha }{1+\cos \alpha }$

Step 2. Concept used.

Divide the provided statement into two halves (LHS and RHS). Simplify each separately and compare the results. If the two outcomes are identical, the supplied statement is an identity.

Step 3. Calculation.

Now, simplify RHS of $\frac{1-\cos \alpha }{\sin \alpha }=\frac{\sin \alpha }{1+\cos \alpha }$:

$\begin{array}{l}\frac{\sin \alpha }{1+\cos \alpha }=\frac{\sin \alpha }{1+\cos \alpha }\times \frac{1-\cos \alpha }{1-\cos \alpha }\\ \frac{\sin \alpha }{1+\cos \alpha }=\frac{\sin \alpha (1-\cos \alpha )}{1-{\cos }^2\alpha }\\ \because 1-{\cos }^2\alpha ={\sin }^2\alpha \\ \frac{\sin \alpha }{1+\cos \alpha }=\frac{\sin \alpha (1-\cos \alpha )}{{\sin }^2\alpha }\\ =\frac{1-\cos \alpha }{\sin \alpha }\end{array}$

LHS of the equation is$\frac{1-\cos \alpha }{\sin \alpha }$

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