A) \[\frac{12}{13}\]
B) 1
C) 0
D) \[\frac{13}{12}\]
Correct Answer: D
Solution :
\[se{{c}^{4}}\theta -ta{{n}^{4}}\theta \] \[=(se{{c}^{2}}\theta +ta{{n}^{2}}\theta )(se{{c}^{2}}\theta -\,ta{{n}^{2}}\theta )\] \[={{\sec }^{2}}\theta +{{\tan }^{2}}\theta =\frac{13}{12}\]
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