A) \[-1\]
B) 0
C) \[se{{c}^{2}}x\]
D) 1
Correct Answer: D
Solution :
(d): \[{{\left( secx.secy+tanx.tany \right)}^{2}}-{{\left( secx.tany+tanx.secy \right)}^{2}}\] \[=se{{c}^{2}}x.se{{c}^{2}}y+ta{{n}^{2}}x.ta{{n}^{2}}y+\]\[2sec\,x\,.\,sec\,y\,.\,tan\,x\,.\,tan\,y-se{{c}^{2}}\] \[ta{{n}^{2}}y-ta{{n}^{2}}x\,.\,se{{c}^{2}}y-2\,sec\,x\,.\,\,secy\,.\,\,tanx\,.\text{ }tany\] \[=se{{c}^{2}}x\,.\,se{{c}^{2}}y-se{{c}^{2}}x\,.\,ta{{n}^{2}}y-ta{{n}^{2}}.\,se{{c}^{2}}y+ta{{n}^{2}}x\,.\,ta{{n}^{2}}y\] \[=se{{c}^{2}}x\left( se{{c}^{2}}y-ta{{n}^{2}}y \right)-ta{{n}^{2}}x\left( se{{c}^{2}}y-ta{{n}^{2}}y \right)\] \[=se{{c}^{2}}x-ta{{n}^{2}}x=1\]
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