| (i) \[3{{(\sin \theta -\cos \theta )}^{4}}+6{{(\sin \theta +\cos \theta )}^{2}}\] \[+4({{\sin }^{6}}\theta +{{\cos }^{6}}\theta )\] is independent of \[\theta \]. |
| (ii) If \[\cos ec\,\theta -\sin \theta ={{a}^{3}},\] \[\sec \theta -\cos \theta ={{b}^{3}},\]then \[{{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})=2\] |
A) Only (i)
B) Only (ii)
C) Both (i) and (ii)
D) Neither (i) or (ii)
Correct Answer: A
Solution :
We know, \[3{{(\sin \theta -\cos \theta )}^{4}}=3{{({{(\sin \theta -\cos \theta )}^{2}})}^{2}}\] \[=3({{1}^{2}}+4{{\sin }^{2}}\theta {{\cos }^{2}}\theta -4\sin \theta \cos \theta )\] ?.(i) and, \[6{{(\sin \theta +\cos \theta )}^{2}}=6+12\sin \theta \cos \theta \] ..?(ii) Also,\[4({{\sin }^{6}}\theta +{{\cos }^{6}}\theta )=4+({{({{\sin }^{2}}\theta )}^{3}}+{{({{\cos }^{2}}\theta )}^{3}})\] \[=4(1)-12{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] ?..(iii) Adding (i), (ii) and (iii), we get \[3+12{{\sin }^{2}}\theta {{\cos }^{2}}\theta -12\sin \theta \cos \theta +6\] \[+12\sin \theta \,\,\cos \theta +4-12\,{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] \[=3+6+4=13\] and 13 is independent of \[\theta \]. (ii) We have, \[\cos ec\,\theta -\sin \theta =\frac{1}{\sin \theta }-\sin \theta =\frac{1-{{\sin }^{2}}\theta }{\sin \theta }\] \[\Rightarrow \] \[{{a}^{3}}=\frac{{{\cos }^{2}}\theta }{\sin \theta }\] or \[{{a}^{2}}={{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{2/3}}\] Similarly, \[\sec \theta -\cos \theta =\frac{1}{\cos \theta }\cos \theta =\frac{1-{{\cos }^{2}}\theta }{\cos \theta }\] \[\Rightarrow \] \[{{b}^{3}}=\frac{{{\sin }^{2}}\theta }{\cos \theta }\] or \[{{b}^{2}}={{\left( \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2/3}}\] \[{{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})={{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta }\times \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2/3}}\] \[\left( {{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{2/3}}+{{\left( \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2/3}} \right)\] \[={{(\sin \theta .\cos \theta )}^{2/3}}\left( \frac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{(\sin \theta .\cos \theta )}^{2/3}}} \right)=1\]
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