A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{4}\]
D) \[2\sqrt{2}\]
Correct Answer: A
Solution :
Given, \[\sin \theta +\cos \theta =\sqrt{2}\] ?.(i) Now, \[{{({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}^{3}}={{\sin }^{6}}\theta +{{\cos }^{6}}\theta \] \[+3{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] \[\Rightarrow \] \[1={{\sin }^{6}}\theta +{{\cos }^{6}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta .1\] \[\Rightarrow \] \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] \[\Rightarrow \]\[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-\frac{3}{4}{{(\sin \theta +\cos \theta )}^{2}}-1{{)}^{2}}\] \[\left[ \begin{align} & \text{since, (sin}\theta \text{+cos}\theta {{\text{)}}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \,\cos \theta \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,2\sin \theta \,\cos \theta ={{(\sin \theta +\cos \theta )}^{2}}-1 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\sin \theta \cos \theta =\frac{1}{2}[{{(\sin \theta +\cos \theta )}^{2}}-1] \\ & \,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,{{\sin }^{2}}\theta {{\cos }^{2}}\theta =\frac{1}{4}{{[{{(\sin \theta +\cos \theta )}^{2}}-1]}^{2}} \\ \end{align} \right]\] \[=1-\frac{3}{4}{{[{{(\sqrt{2})}^{2}}-1]}^{2}}\] [from Eq. (i)] \[\Rightarrow \] \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-\frac{3}{4}{{[2-1]}^{2}}\] \[=1-\frac{3}{4}\] \[\Rightarrow \] \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =\frac{1}{4}\]
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