Problem

Write each expression in terms of sine and cosine, and then simplify so that no quotients appear in the final expression and all functions are of $\theta$ only.
\[
\sin ^{2}(-\theta)+\cot ^{2}(-\theta)+\cos ^{2}(-\theta)
\]
\[
\sin ^{2}(-\theta)+\cot ^{2}(-\theta)+\cos ^{2}(-\theta)=
\]

Answer

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Answer

Since \(\sin^2\theta\) and \(\cos^2\theta\) are both less than or equal to 1, the final expression is always less than or equal to 2. Therefore, the simplified expression is \(\boxed{1 + \cos^2\theta\sin^2\theta}\).

Steps

Step 1 :First, we write each function in terms of sine and cosine. We have \(\sin^2(-\theta) = \sin^2\theta\), \(\cos^2(-\theta) = \cos^2\theta\), and \(\cot^2(-\theta) = \cot^2\theta\).

Step 2 :Next, we express \(\cot^2\theta\) in terms of sine and cosine. We know that \(\cot\theta = \frac{\cos\theta}{\sin\theta}\), so \(\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta}\).

Step 3 :Substituting these into the original expression, we get \(\sin^2\theta + \frac{\cos^2\theta}{\sin^2\theta} + \cos^2\theta\).

Step 4 :To simplify this expression, we multiply the second term by \(\sin^2\theta\) to get \(\sin^2\theta + \cos^2\theta + \cos^2\theta\sin^2\theta\).

Step 5 :Finally, we use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) to simplify the expression to \(1 + \cos^2\theta\sin^2\theta\).

Step 6 :Since \(\sin^2\theta\) and \(\cos^2\theta\) are both less than or equal to 1, the final expression is always less than or equal to 2. Therefore, the simplified expression is \(\boxed{1 + \cos^2\theta\sin^2\theta}\).

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