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 \[se{{c}^{2}}\theta -ta{{n}^{2}}\theta =1\] \[\therefore se{{c}^{4}}\theta -ta{{n}^{4}}\theta \] \[=\left( se{{c}^{2}}\theta +ta{{n}^{2}}\theta \right)\left( se{{c}^{2}}\theta -ta{{n}^{2}}\theta \right)\] \[=\sqrt{2}\times 1=\sqrt{2}\]You need to login to perform this action.
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