A) \[{{\sin }^{-1}}\,\left[ \frac{n-1}{n+1} \right]\]
B) \[{{\sin }^{-1}}\,\left[ \frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right]\]
C) \[{{\cos }^{-1}}\,\left[ \frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right]\]
D) \[{{\cos }^{-1}}\,\left[ \frac{n-1}{n+1} \right]\]
Correct Answer: C
Solution :
\[\left| \overrightarrow{A}+\overrightarrow{B}\text{ } \right|=n\left| \overrightarrow{A}-\overrightarrow{B}\text{ } \right|\] \[{{A}^{2}}+{{B}^{2}}+2\,AB\text{ }cos\text{ }\theta ={{n}^{2}}\left( {{A}^{2}}+{{B}^{2}}-2\,AB\text{ }cos\text{ }\theta \right)\]\[2AB\text{ }cos\text{ }\theta \left( 1+{{n}^{2}} \right)=\left( {{A}^{2}}+{{B}^{2}} \right)\left( {{n}^{2}}-1 \right)\] \[\cos \,\theta \,\,=\,\frac{{{A}^{2}}+{{B}^{2}}}{2\,AB}\,\frac{({{n}^{2}}-1)}{({{n}^{2}}+1)}\] As \[A=B\] \[\therefore \,\,\cos \,\theta \,=\frac{{{B}^{2}}+{{B}^{2}}}{2{{B}^{2}}}\,\left( \frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right)\] \[\cos \,\theta \,\,=\,\,\frac{{{n}^{2}}-1}{{{n}^{2}}+1}\] \[\,\theta \,\,=\,\,{{\cos }^{-1}}\,\left( \,\frac{{{n}^{2}}-1}{{{n}^{2}}+1} \right)\]
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