verify the Pythagorean identity. 1+\cot ^2\theta \csc ^2\theta
Starting with the left-hand side:
\begin{align*}
\text{LHS} &= \frac{1}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \\
&= \frac{1 + \cos^2\theta}{\sin^2\theta} \\
&= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta} \\
&= \frac{1}{\sin^2\theta}
\end{align*}
Now, we use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ to simplify the right-hand side:
\begin{align*}
\text{RHS} &= \frac{1}{\sin^2\theta} \\
&= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta} \\
&= \frac{ \cos^2\theta}{\sin^2\theta} \cdot \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} \\
&= \frac{1}{\cos^2\theta} \cdot \frac{ \cos^2\theta}{\sin^2\theta} + \frac{1}{\sin^2\theta} \\
&= \cot^2\theta \csc^2\theta + 1
\end{align*}
Therefore, we have shown that the identity
$$1 + \cot^2\theta \csc^2\theta = \frac{1}{\sin^2\theta}$$
is true.
To verify the Pythagorean identity, we need to show that
1 + cot^2(theta) = csc^2(theta)
We'll start by expressing the cotangent and cosecant functions in terms of sine and cosine:
cot(theta) = cos(theta) / sin(theta)
csc(theta) = 1 / sin(theta)
Now, let's rewrite the left side of the equation using these expressions:
1 + cot^2(theta) = 1 + (cos(theta) / sin(theta))^2
Next, let's simplify the expression on the right side of the equation:
= 1 + cos^2(theta) / sin^2(theta)
To get a common denominator, we multiply the numerator and denominator of 1 by sin^2(theta):
= (sin^2(theta) / sin^2(theta)) + cos^2(theta) / sin^2(theta)
= (sin^2(theta) + cos^2(theta)) / sin^2(theta)
Since sin^2(theta) + cos^2(theta) is equal to 1 (which is a fundamental trigonometric identity), we can substitute this value:
= 1 / sin^2(theta)
This matches the right side of the equation, so we have verified the Pythagorean identity:
1 + cot^2(theta) = csc^2(theta)