A) \[\frac{1}{\sqrt{3}}\]
B) 1
C) \[\sqrt{2}\]
D) 0
Correct Answer: C
Solution :
(c): \[se{{c}^{2}}\theta +ta{{n}^{2}}\theta =\sqrt{2}\] And \[se{{c}^{2}}\theta -{{\tan }^{2}}\theta =1\] \[\therefore se{{c}^{4}}\theta -ta{{n}^{4}}\theta \] = (sec2 9 + tan2 9)(sec2 9 - tan2 9) \[=\sqrt{2}+1=\sqrt{2}\]You need to login to perform this action.
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