A) \[4{{\sin }^{2}}\alpha \]
B) \[-4{{\sin }^{2}}\alpha \]
C) \[2\sin 2\alpha \]
D) \[4\]
Correct Answer: A
Solution :
If \[{{\cos }^{-1}}\frac{x}{a}+{{\cos }^{-1}}\frac{y}{b}=\theta \] Then \[\frac{{{x}^{2}}}{{{a}^{2}}}\cos \theta +\frac{{{y}^{2}}}{{{b}^{2}}}={{\sin }^{2}}\theta \] Here \[{{\cos }^{-1}}\frac{x}{1}+{{\cos }^{-1}}\frac{y}{2}=\alpha ;\] \[\therefore \frac{{{x}^{2}}}{1}-\frac{2xy}{2}\cos \alpha +\frac{{{y}^{2}}}{4}={{\sin }^{2}}\alpha \] \[{{x}^{2}}-xy\cos \alpha +\frac{{{y}^{2}}}{4}={{\sin }^{2}}\alpha \] \[4{{x}^{2}}-4xy\cos \alpha +{{y}^{2}}=4{{\sin }^{2}}\alpha \].
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