A) 0
B) 1
C) 2
D) None of these
Correct Answer: B
Solution :
| Given, \[\alpha +\beta =90{}^\circ \] ?(i) |
| \[\therefore \] \[\sqrt{(\text{cosec}\alpha \cdot \text{cosec}\beta \text{)}}{{\left( \frac{\sin \alpha }{\sin \beta }+\frac{\cos \alpha }{\cos \beta } \right)}^{1/2}}\] |
| \[=\frac{1}{{{(\sin \alpha \sin \beta )}^{1/2}}}{{\left( \frac{\sin \alpha \cos \beta +\cos \alpha +\sin \beta }{\sin \beta \cos \beta } \right)}^{-1/2}}\] |
| \[=\frac{1}{{{(sin\alpha sin\beta )}^{1/2}}}{{\left\{ \frac{\sin (\alpha +\beta )}{\sin \beta \cos \beta } \right\}}^{-1/2}}\] |
| \[=\frac{1}{{{(\sin \alpha \sin \beta )}^{1/2}}}{{\left\{ \frac{\sin 90{}^\circ }{\cos (90-\alpha )\sin \beta } \right\}}^{-1/2}}\] |
| [from Eq. (i)] |
| \[=\frac{1}{{{(\sin \alpha \sin \beta )}^{1/2}}}\times (\sin \alpha \sin \beta )1/2=1\] |
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