verify the Pythagorean identity. 1+\cot ^2\theta \csc ^2\theta

Answer 1

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.