Question

Use trigonometric identities to verify each expression is equal. sin(x)/1-cos(x)-cot(x)=csc(x)

Answer 1

Answer:

See Below.

Step-by-step explanation:

We want to verify the identity:

$\displaystyle \frac{\sin x}{1 - \cos x} - \cot x = \csc x$

We can multiply the fraction by 1 + cos(x):

$\displaystyle \frac{\sin x(1+\cos x)}{(1 - \cos x)(1+\cos x)} - \cot x = \csc x$

Difference of Two Squares:

$\displaystyle \frac{\sin x(1+\cos x)}{1-\cos^2 x} - \cot x = \csc x$

From the Pythagorean Theorem, we know that sin²(x) + cos²(x) = 1. Rearranging, we acquire that sin²(x) = 1 - cos²(x). Substitute:

$\displaystyle \frac{\sin x(1+\cos x)}{\sin^2 x} - \cot x = \csc x$

Cancel:

$\displaystyle \frac{ 1 + \cos x}{\sin x}-\cot x = \csc x$

Let cot(x) = cos(x) / sin(x):

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

Combine Fractions:

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

Thus:

$\displaystyle \frac{1}{\sin x}=\csc x = \csc x$

Hence proven.