A) \[-\,\,1\]
B) \[1\]
C) \[2\]
D) \[-\,\,2\]
Correct Answer: B
Solution :
| \[\text{cosec}\theta -\sin \theta ={{a}^{3}}\] |
| \[\frac{1-{{\sin }^{2}}\theta }{\sin \theta }={{a}^{3}}\] |
| \[\Rightarrow \] \[\frac{{{\cos }^{2}}\theta }{\sin \theta }={{a}^{3}}\] |
| \[{{\cos }^{2}}\theta ={{a}^{3}}\sin \theta \] ?(i) |
| \[\sec \theta -\cos \theta ={{b}^{3}}\] |
| \[\frac{1-{{\cos }^{2}}\theta }{\cos \theta }={{b}^{3}}\] |
| \[\Rightarrow \] \[\frac{{{\sin }^{2}}\theta }{\cos \theta }={{b}^{3}}\] |
| \[{{\sin }^{2}}\theta ={{b}^{3}}\cos \theta \] ?(ii) |
| Put in Eq. (i), \[\cos \theta =\frac{{{\sin }^{2}}\theta }{{{b}^{3}}}\] |
| \[\frac{{{\sin }^{4}}\theta }{{{b}^{6}}}={{a}^{3}}\sin \theta \] |
| \[\Rightarrow \] \[{{\sin }^{3}}\theta ={{a}^{2}}{{b}^{6}}\] |
| \[\therefore \] \[\sin \theta =a{{b}^{2}}\] ?(iii) |
| Similarly, \[\cos \theta ={{a}^{2}}b\] ?(iv) |
| Squaring and adding Eqs. (iii) and (iv), we get |
| We have,\[1={{a}^{2}}{{b}^{4}}+{{a}^{4}}{{b}^{2}};\]\[1={{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})\] |
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