Question

True or false? cscx+sinx
=
cot^2x⋅sinx
is an identity.



True

False

Answer 1

To determine whether the equation \( \csc x + \sin x = \cot^2 x \cdot \sin x \) is an identity, we will simplify both sides.

Starting with the left side:

\[
\csc x + \sin x = \frac{1}{\sin x} + \sin x
\]
To combine these terms, we can express \( \sin x \) with a common denominator:
\[
= \frac{1}{\sin x} + \frac{\sin^2 x}{\sin x} = \frac{1 + \sin^2 x}{\sin x}
\]

Next, we simplify the right side:

\[
\cot^2 x \cdot \sin x
\]
Recall that \( \cot x = \frac{\cos x}{\sin x} \), so:
\[
\cot^2 x = \frac{\cos^2 x}{\sin^2 x}
\]
Thus,
\[
\cot^2 x \cdot \sin x = \frac{\cos^2 x}{\sin^2 x} \cdot \sin x = \frac{\cos^2 x}{\sin}
\]

Now we can rewrite the identity question:
\[
\frac{1 + \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}
\]

Eliminating the common denominator \( \sin x \) (assuming \( \sin x \neq 0 \)):
\[
1 + \sin^2 x = \cos^2 x
\]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[
\cos^2 x = 1 - \sin^2 x
\]

Setting the two right-hand sides equal gives:
\[
1 + \sin^2 x = 1 - \sin^2 x
\]
\[
\sin^2 x + \sin^2 x = 0
\]
\[
2\sin^2 x = 0
\]
So \( \sin^2 x = 0 \) implies \( \sin x = 0 \), which is true for specific values of \( x \).

Since both sides are only equal under a specific condition (i.e., \( \sin x = 0 \)) and not universally true for all \( x \), we can conclude:

**False**: The equation \( \csc x + \sin x = \cot^2 x \cdot \sin x \) is not an identity.