A) \[1-{{a}^{2}}-{{b}^{2}}\]
B) \[1-2{{a}^{2}}-2{{b}^{2}}\]
C) \[2+{{a}^{2}}+{{b}^{2}}\]
D) \[2-{{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
Given that \[\sin \,(\theta +\alpha )=a\] ?..(i) and \[\sin \,(\theta +\beta )=b\] ?..(ii) Now, \[\cos \,(\theta +\alpha )=\sqrt{1-{{a}^{2}}}\,\Rightarrow \,\,\theta +\alpha ={{\cos }^{-1}}\sqrt{1-{{a}^{2}}}\] and \[\alpha \,-\beta =(\theta +\alpha )-(\theta +\beta )\] \[=\,\,{{\cos }^{-1}}\sqrt{1-{{a}^{2}}}-{{\cos }^{-1}}\sqrt{1-{{b}^{2}}}\] \[\Rightarrow \,\,\alpha -\beta ={{\cos }^{-1}}(\sqrt{1-{{a}^{2}}}\,\sqrt{1-{{b}^{2}}}+ab)\] \[\Rightarrow \,\,\cos \,(\alpha -\beta )=\sqrt{1-{{a}^{2}}}\,\sqrt{1-{{b}^{2}}}+ab\] Now, \[\cos \,\,2\,(\alpha -\beta )-4ab\,\,\cos \,(\alpha -\beta )\] \[=2\,\,{{\cos }^{2}}\,(\alpha -\beta )-1-4ab\,\,\cos \,(\alpha -\beta )\] \[=2\,{{\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)}^{2}}\] \[-4ab\,\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)-1\] \[=2\,\{(1-{{a}^{2}})(1-{{b}^{2}})+{{a}^{2}}{{b}^{2}}+2ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}\}\] \[-4ab\,(\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab)\] \[=\,\,2\,(1-{{b}^{2}}-{{a}^{2}}+{{a}^{2}}{{b}^{2}})+2{{a}^{2}}{{b}^{2}}-4{{a}^{2}}{{b}^{2}}-1\] \[=\,\,2\,(1-{{a}^{2}}-{{b}^{2}})-1=1-2{{a}^{2}}-2{{b}^{2}}.\]
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