A) \[\frac{({{a}^{2}}+{{b}^{2}})}{({{a}^{2}}-{{b}^{2}})}\]
B) \[\frac{({{a}^{2}}-{{b}^{2}})}{({{a}^{2}}+{{b}^{2}})}\]
C) \[\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})}\]
D) \[\frac{{{b}^{2}}}{({{a}^{2}}+{{b}^{2}})}\]
Correct Answer: B
Solution :
| [b] We have \[\tan \theta =\frac{a}{b}\] |
| Now,\[\frac{(a\sin \theta -b\cos \theta )}{(a\sin \theta +b\cos \theta )}=\frac{(a\tan \theta -b)}{(a\tan \theta +b)}\] |
| (Dividing numerator and denominator by \[\cos \theta \]] |
| \[=\frac{\left( a\times \frac{a}{b}-b \right)}{\left( a\times \frac{a}{b}+b \right)}=\frac{({{a}^{2}}-{{b}^{2}})}{({{a}^{2}}+{{b}^{2}})}\] |
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