Question:
$\sec^{2}x + \csc^{2}x - \sec^{2}x \csc^{2}x =$
$\sec^{2}x + \csc^{2}x - \sec^{2}x \csc^{2}x =$
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The identity $\sec^2 \theta + \csc^2 \theta = \sec^2 \theta \csc^2 \theta$ is a very useful shortcut in trigonometry.
Updated On: Apr 28, 2026
- $\sec^{2}x$
- $\csc^{2}x$
- $\cot^{2}x$
- 1
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Solution and Explanation
Step 1: Concept
Convert the trigonometric functions into sine and cosine.
Step 2: Analysis
$\sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.
Step 3: Calculation
The expression becomes $\frac{1}{\sin^2 x \cos^2 x} - \sec^2 x \csc^2 x$. Since $\sec^2 x \csc^2 x = \frac{1}{\cos^2 x} \cdot \frac{1}{\sin^2 x}$, the expression is $\frac{1}{\sin^2 x \cos^2 x} - \frac{1}{\sin^2 x \cos^2 x} = 0$. Final Answer: (E)
Convert the trigonometric functions into sine and cosine.
Step 2: Analysis
$\sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.
Step 3: Calculation
The expression becomes $\frac{1}{\sin^2 x \cos^2 x} - \sec^2 x \csc^2 x$. Since $\sec^2 x \csc^2 x = \frac{1}{\cos^2 x} \cdot \frac{1}{\sin^2 x}$, the expression is $\frac{1}{\sin^2 x \cos^2 x} - \frac{1}{\sin^2 x \cos^2 x} = 0$. Final Answer: (E)
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