Sure! Let's solve the given trigonometric equation step-by-step:
Given: [tex]\(\frac{\csc ^2 \theta}{\cot \theta} = \csc \theta \sec \theta\)[/tex]
To prove this identity is true, we need to recall and use certain trigonometric identities:
1. [tex]\(\csc \theta = \frac{1}{\sin \theta}\)[/tex]
2. [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex]
3. [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]
4. [tex]\(\csc^2 \theta = \left(\frac{1}{\sin \theta}\right)^2 = \frac{1}{\sin^2 \theta}\)[/tex]
We will substitute these identities into the given equation.
Left-hand side (LHS):
[tex]\[
\frac{\csc ^2 \theta}{\cot \theta} = \frac{\frac{1}{\sin^2 \theta}}{\frac{\cos \theta}{\sin \theta}}
\][/tex]
To simplify this, divide the numerator by the denominator:
[tex]\[
\frac{\frac{1}{\sin^2 \theta}}{\frac{\cos \theta}{\sin \theta}} = \frac{1}{\sin^2 \theta} \times \frac{\sin \theta}{\cos \theta} = \frac{1 \times \sin \theta}{\sin^2 \theta \times \cos \theta} = \frac{\sin \theta}{\sin^2 \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
\][/tex]
Right-hand side (RHS):
[tex]\[
\csc \theta \sec \theta = \left(\frac{1}{\sin \theta}\right) \left(\frac{1}{\cos \theta}\right) = \frac{1}{\sin \theta \cos \theta}
\][/tex]
Comparison:
We see that the simplified LHS is:
[tex]\[
\frac{1}{\sin \theta \cos \theta}
\][/tex]
And the RHS is also:
[tex]\[
\frac{1}{\sin \theta \cos \theta}
\][/tex]
Since the LHS equals the RHS, the identity holds true.
Thus, the trigonometric identity [tex]\(\frac{\csc ^2 \theta}{\cot \theta} = \csc \theta \sec \theta\)[/tex] is indeed correct.
