A) \[\frac{19}{25}\]
B) \[\frac{18}{35}\]
C) \[\frac{17}{25}\]
D) \[\frac{18}{25}\]
Correct Answer: C
Solution :
\[\cot \theta =\frac{7}{24},\,\,\,\,\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] \[=1+\frac{49}{576}=\frac{625}{576}={{\left( \frac{25}{24} \right)}^{2}}\] \[\therefore \] \[\cos ec\,\,\theta =\pm \frac{25}{24}\] \[\therefore \,\,\]\[\sin \theta =\pm \frac{24}{25}\] \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta =1-\frac{576}{625}=\frac{49}{625}\] \[\cos \theta =\pm \frac{7}{25}\,\,\,\,\,\,\,\text{as}\,\,\,\,\,\pi <\theta <\frac{3\pi }{2}\] \[\therefore \sin \theta \,\]and\[\cos \theta \]both are negative. \[\therefore \]\[\sin \theta =\frac{-24}{25},\cos \theta =\frac{-7}{25}\] \[\therefore \,\,\]\[\cos \theta -\sin \theta =\frac{-7}{25}+\frac{24}{25}=\frac{17}{25}\]
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