A) \[{{\cos }^{-1}}\left[ -\frac{({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})} \right]\]
B) \[{{\cos }^{-1}}\left[ \frac{-2({{x}^{2}}-{{y}^{2}})}{({{x}^{2}}+{{y}^{2}})} \right]\]
C) \[{{\cos }^{-1}}\left[ -\frac{({{x}^{2}}+{{y}^{2}})}{({{x}^{2}}-{{y}^{2}})} \right]\]
D) \[{{\cos }^{-1}}\left[ -\frac{({{x}^{2}}-{{y}^{2}})}{({{x}^{2}}+{{y}^{2}})} \right]\]
Correct Answer: A
Solution :
We know that \[{{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] On substituting the value \[A=(x+y),\,B=(x-y)\] \[{{x}^{2}}+{{y}^{2}}\,={{(x+y)}^{2}}\,+{{(x-y)}^{2}}+2(x-y)\,\cos \theta \] \[\Rightarrow \,{{x}^{2}}+{{y}^{2}}\,={{x}^{2}}+{{y}^{2}}+2xy\,+{{x}^{2}}+{{y}^{2}}\,-2xy\] \[+2({{x}^{2}}-{{y}^{2}})\,\cos \theta \] \[\Rightarrow \,\,\frac{-({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})}\,=\cos \theta \] \[\Rightarrow \,\theta =\,{{\cos }^{-1}}\,\left[ -\,\frac{({{x}^{2}}+{{y}^{2}})}{2({{x}^{2}}-{{y}^{2}})} \right]\]
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