A) \[0\]
B) \[1\]
C) \[-1\]
D) \[\pm 1\]
Correct Answer: D
Solution :
| [d]\[x=(\sec A-\tan A)\,\,(\sec B-\tan B)\,\,(\sec C-\tan C)\] |
| \[=(\sec A+\tan A)\,\,(\sec B+\tan B)(\sec C+\tan C)\] |
| \[\Rightarrow \,\,\,(\sec A+tanA)(\sec B+\tan B)(\sec C+\tan C)\] |
| \[(\sec A-\tan A)(\sec B-\tan B)(\sec C-\tan C)\] |
| \[={{\{(\sec A+\tan A)(\sec B+\tan B)(\sec A+\tan A)\}}^{2}}\]\[\Rightarrow \,\,\,({{\sec }^{2}}A-{{\tan }^{2}}A)({{\sec }^{2}}B-{{\tan }^{2}}B)({{\sec }^{2}}C-{{\tan }^{2}}C)={{x}^{2}}\]\[\Rightarrow \,\,\,{{x}^{2}}=1\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,x=\pm 1\] |
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