A) \[\upsilon /\text{1}0\]
B) \[f\]
C) \[1.11f\]
D) \[1.22f\]
Correct Answer: A
Solution :
Now \[\alpha -\beta =(\theta -\beta )-(\theta -\alpha )\] \[\therefore \] \[\cos (\alpha -\beta )=\cos (\theta -\beta )\cos (\theta -\alpha )\] \[+sin(\theta -\beta )\sin (\theta -\alpha )\] \[=ab+\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}\] And \[\sin (\alpha -\beta )=\pm (a\sqrt{1-{{b}^{2}}}-b\sqrt{1-{{a}^{2}}})\] \[\Rightarrow \] \[{{\sin }^{2}}(\alpha -\beta )={{a}^{2}}+{{b}^{2}}-2{{a}^{2}}{{b}^{2}}\] \[-2ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}\] \[\Rightarrow \] \[{{\sin }^{2}}(\alpha -\beta )={{a}^{2}}+{{b}^{2}}-2{{a}^{2}}{{b}^{2}}\] \[-2ab[\cos (\alpha -\beta )-ab]\] \[\Rightarrow \] \[{{\sin }^{2}}(\alpha -\beta )={{a}^{2}}+{{b}^{2}}-2ab\,\cos (\alpha -\beta )\] \[\Rightarrow \] \[{{\sin }^{2}}(\alpha -\beta )+2ab\cos (\alpha -\beta )={{a}^{2}}+{{b}^{2}}\]
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