Question

Proving Identities

Verify the identity$\frac{\cos \theta }{1-\sin \theta }=\frac{\sin \theta -\csc \theta }{\cos \theta -\cot \theta }$

Short answer

The expression$\frac{\cos \theta }{1-\sin \theta }=\frac{\sin \theta -\csc \theta }{\cos \theta -\cot \theta }$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{\cos \theta }{1-\sin \theta }=\frac{\sin {\displaystyle }\theta -\csc \theta }{\cos {\displaystyle }\theta -\cot \theta }$

Step 2. Concept used.

Divide the given statement into two pieces, LHS and RHS, to demonstrate that it is an identity. After you've finished separating, separate the two and simplify them separately. Given expression is an identity if both outcomes are equivalent.

Step 3. Calculation.

Now, simplify RHS of $\frac{\cos \theta }{1-\sin \theta }=\frac{\sin \theta -\csc \theta }{\cos \theta -\cot \theta }$:

$\begin{array}{l}\frac{\sin \theta -\csc \theta }{\cos \theta -\cot \theta }=\frac{\sin \theta -\frac{1}{\sin \theta }}{\cos \theta -\frac{\cos \theta }{\sin \theta }}\\ =\frac{{\sin }^2\theta -1}{\cos \theta \sin \theta -\cos \theta }\times \frac{\sin \theta }{\sin \theta }\end{array}$

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

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

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