Question

True or false? [(1−cost)/sint]+[sint/(1−cost)]
= 2csct
is an identity.




True

False

Answer 1

To determine if the equation

\[
\frac{1 - \cos t}{\sin t} + \frac{\sin t}{1 - \cos t} = 2 \csc t
\]

is an identity, we can simplify the left-hand side and see if it simplifies to the right-hand side.

Starting with the left-hand side:

1. Rewrite the expression:

\[
LHS = \frac{1 - \cos t}{\sin t} + \frac{\sin t}{1 - \cos t}
\]

2. Find a common denominator for the two fractions. The common denominator is \(\sin t (1 - \cos t)\):

\[
LHS = \frac{(1 - \cos t)^2 + \sin^2 t}{\sin t (1 - \cos t)}
\]

3. Now simplify the numerator:

\[
(1 - \cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t
\]

Using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\):

\[
= 1 - 2\cos t + 1 = 2 - 2\cos t = 2(1 - \cos t)
\]

4. Substituting this back into the left-hand side:

\[
LHS = \frac{2(1 - \cos t)}{\sin t (1 - \cos t)}
\]

Assuming \(1 - \cos t \neq 0\) (or when \(t \neq 2n\pi\) for some integer \(n\)), we can cancel \(1 - \cos t\):

\[
LHS = \frac{2}{\sin t} = 2 \csc t
\]

Now we have shown that:

\[
LHS = 2 \csc t
\]

Thus,

\[
\frac{1 - \cos t}{\sin t} + \frac{\sin t}{1 - \cos t} = 2 \csc t
\]

is indeed an identity.

Therefore, the answer is **True**.