| If \[({{a}^{2}}-{{b}^{2}})\sin \theta +2ab\cos \theta ={{a}^{2}}+{{b}^{2}},\]then the value of tan \[\theta \] is |
A) \[\frac{1}{2ab}\,({{a}^{2}}+{{b}^{2}})\]
B) \[\frac{1}{2}({{a}^{2}}-{{b}^{2}})\]
C) \[\frac{1}{2ab}\,({{a}^{2}}-{{b}^{2}})\]
D) \[\frac{1}{2}({{a}^{2}}+{{b}^{2}})\]
Correct Answer: C
Solution :
| \[({{a}^{2}}-{{b}^{2}})\sin \theta +2ab\cos \theta ={{a}^{2}}+{{b}^{2}}\] |
| \[\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)\sin \theta +\frac{2ab}{{{a}^{2}}+{{b}^{2}}}\cos \theta =1\] |
 |
| \[\sin \alpha =\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] |
| \[\cos \alpha =\frac{2ab}{{{a}^{2}}+{{b}^{2}}}\] |
| \[\therefore \] \[\sin \alpha sin\theta +\cos \alpha \cos \theta =1\] |
| \[\Rightarrow \] \[\cos \,(\alpha -\theta )=1=cos\,(0{}^\circ )\] |
| \[\Rightarrow \] \[\alpha -\theta =0{}^\circ \]\[\Rightarrow \]\[\alpha =\theta \] |
| \[\Rightarrow \] \[\tan \alpha =\tan \theta =\frac{{{a}^{2}}-{{b}^{2}}}{2ab}\] |
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