| If \[\sin \theta +\cos \theta =\frac{17}{13},\]\[0{}^\circ <\theta <90{}^\circ ,\]then the value of \[\sin \theta -\cos \theta \] is [SSC (CGL) 2011] |
A) \[\frac{5}{17}\]
B) \[\frac{3}{19}\]
C) \[\frac{7}{10}\]
D) \[\frac{7}{13}\]
Correct Answer: D
Solution :
| \[\sin \theta +\cos \theta =\frac{17}{13}\] |
| On squaring both sides, we get |
| \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{\left( \frac{17}{13} \right)}^{2}}\] |
| \[\Rightarrow \] \[2\sin \theta \cos \theta ={{\left( \frac{17}{13} \right)}^{2}}-1\] |
| \[\Rightarrow \] \[2\sin \theta \cos \theta =\left( \frac{17}{13}-1 \right)\,\,\left( \frac{17}{13}+1 \right)\] |
| \[=\left( \frac{4}{13} \right)\,\,\left( \frac{30}{13} \right)=\frac{120}{169}\] |
| Now, \[{{(\sin \theta -\cos \theta )}^{2}}={{(\sin \theta +\cos \theta )}^{2}}-4\sin \theta \cos \theta \] |
| \[={{\left( \frac{17}{13} \right)}^{2}}-2\times \frac{120}{169}\] |
| \[=\frac{289-240}{169}=\frac{49}{169}\] |
| \[\therefore \] \[\sin \theta -\cos \theta =\frac{7}{13}\] |
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