Discussion
\(sec^{2}x + cosec^{2}x - sec^{2}xcosec^{2}x =\)
Answer: E
The value is 0.
Add the first two terms over a common denominator.
\[ \begin{aligned} sec^{2}x + cosec^{2}x &= \frac{1}{\cos^{2}x} + \frac{1}{\sin^{2}x} \\ &= \frac{\sin^{2}x + \cos^{2}x}{\sin^{2}x\cos^{2}x} = sec^{2}x\,cosec^{2}x \end{aligned} \]
So the sum of the squares equals their product and the expression is 0 for every x where it is defined. The answers 1 or a single trigonometric function come from stopping before using \(\sin^{2}x + \cos^{2}x = 1\).
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