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