| If\[\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{5}{4},\] then the value of \[\frac{{{\tan }^{2}}\theta +1}{{{\tan }^{2}}\theta -1}\] is [SSC(10+2)2012] |
A) \[\frac{25}{16}\]
B) \[\frac{41}{9}\]
C) \[\frac{41}{40}\]
D) \[\frac{40}{41}\]
Correct Answer: C
Solution :
| Given,\[\frac{\sin \theta +cos\theta }{\sin \theta -\cos \theta }=\frac{5}{4}\] |
| On dividing by \[\cos \theta \]in numerator and denominator respectively. |
| \[\Rightarrow \] \[\frac{\frac{\sin \theta }{\cos \theta }+1}{\frac{\sin \theta }{\cos \theta }-1}=\frac{5}{4}\] |
| \[\Rightarrow \] \[\frac{\tan \theta +1}{\tan \theta -1}=\frac{5}{4}\] |
| \[\Rightarrow \]\[4\tan \theta +4=5\tan \theta -5\] |
| \[\therefore \]\[\frac{{{\tan }^{2}}\theta +1}{{{\tan }^{2}}\theta -1}=\frac{{{9}^{2}}+1}{{{9}^{2}}-1}=\frac{82}{80}=\frac{41}{40}\] |
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