A) 5
B) 4
C) 3
D) None of these
Correct Answer: B
Solution :
Given, \[3\sin \theta +5\cos \theta =5\] \[\Rightarrow \] \[3\sin \theta =5\,(1-cos\theta )\] \[\Rightarrow \] \[3\cdot 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}=5\cdot 2{{\sin }^{2}}\frac{\theta }{2}\] \[\left[ \begin{align} & \because \,\,\sin \theta =2sin\frac{\theta }{2}\cos \frac{\theta }{2} \\ & \text{and}\,\,1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} \\ \end{align} \right]\] \[\Rightarrow \] \[\tan \frac{\theta }{2}=\frac{3}{5}\] Now, \[5\sin \theta -3\cos \theta \] \[=5\cdot \frac{2\tan \frac{\theta }{2}}{1+{{\tan }^{2}}\frac{\theta }{2}}-3\cdot \frac{1-{{\tan }^{2}}\frac{\theta }{2}}{1+{{\tan }^{2}}\frac{\theta }{2}}\] \[=5\cdot \frac{2\cdot \frac{3}{5}}{\left( 1+\frac{9}{25} \right)}-3\cdot \frac{\left( 1-\frac{9}{25} \right)}{\left( 1+\frac{9}{25} \right)}\] \[=\frac{6-3\cdot \frac{16}{25}}{1+\frac{9}{25}}\] \[=\frac{150-48}{34}=\frac{102}{34}=3\]
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