A) 4
B) 1
C) 2
D) 5
Correct Answer: A
Solution :
(a): \[2\beta sin\theta =\alpha cos\theta \] \[\Rightarrow \alpha =\frac{2\beta sin\theta }{cos\theta }\] ??.. (i) \[\therefore 2\alpha cosec\theta -\beta sec\theta =3\] \[\Rightarrow \frac{2\times 2\times \beta sin\theta .cosec\theta }{\cos \theta }-\beta sec\theta =3\] \[\Rightarrow 4\beta sec\theta -\beta sec\theta =3\] \[\Rightarrow 3\beta sec\theta =3\] From equation (i) \[\alpha =\frac{2\times cos\theta .sin\theta }{\cos \theta }=2\sin \theta \] \[\therefore {{\alpha }^{2}}+4{{\beta }^{2}}={{\left( 2\sin \theta \right)}^{2}}+4{{\cos }^{2}}\theta \] \[=4\left( si{{n}^{2}}\theta +co{{s}^{2}}\theta \right)=4\]
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