A) \[cos\theta =x+\frac{1}{x}\]
B) \[sec\theta =\frac{{{x}^{2}}}{1+{{x}^{2}}}\]
C) \[cosec\theta =\frac{{{x}^{2}}}{1+{{x}^{2}}}\]
D) \[\tan \theta =\frac{{{x}^{2}}+x+1}{{{x}^{2}}-3x+5}\]
Correct Answer: D
Solution :
(d): Put \[x=1\] and verify in all the options. Since \[\cos \theta \le 1\], but \[x+\frac{1}{x}\cancel{\le }1\]always. \[\therefore \] Option (a) is not true. Further, \[\sec \theta \ge 1\] but \[\frac{{{x}^{2}}}{1+{{x}^{2}}}<1\] always. \[\therefore \] Option (b) is not possible. Similarly, since \[cosec\theta \ge 1\]always. \[\therefore \] Options (c) is not possible. Hence, by elimination method, option (d) is true. Aliter: You can also verify for option (d) separately as \[-\infty <\tan \theta <+\infty \]
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