A) \[\frac{5}{17}\]
B) \[\frac{3}{19}\]
C) \[\frac{7}{10}\]
D) \[\frac{7}{13}\]
Correct Answer: D
Solution :
| [d] \[\sin \theta +\cos \theta =\frac{17}{13}\] \[\Rightarrow \] \[{{(sin\theta +cos\theta )}^{2}}=\frac{{{(17)}^{2}}}{{{(13)}^{2}}}\] \[\Rightarrow \] \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta \] \[=\frac{289}{169}\] (i) \[\Rightarrow \] \[1+2\sin \theta \cos \theta =\frac{289}{169}\] \[\Rightarrow \] \[2\sin \theta \cos \theta =\frac{120}{169}\] Again, from Eq. (i), we get \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta \] \[+\,4\sin \theta \cos \theta =\frac{289}{169}\] \[\Rightarrow \]\[{{(sin\theta -cos\theta )}^{2}}+4\sin \theta \cos \theta =\frac{289}{169}\] \[\Rightarrow \]\[{{(sin\theta -cos\theta )}^{2}}+2\times \frac{120}{169}=\frac{289}{169}\] [using Eq. (ii)] \[\Rightarrow \]\[{{(sin\theta -cos\theta )}^{2}}=\frac{289}{169}-\frac{240}{169}=\frac{49}{169}\] \[\Rightarrow \]\[\sin \theta -\cos \theta =\frac{7}{13}\] |
You need to login to perform this action.
You will be redirected in
3 sec
