| If \[\cos x+{{\cos }^{2}}x=1,\]then the numerical value of \[({{\sin }^{12}}x+3{{\sin }^{10}}x+3{{\sin }^{8}}x+{{\sin }^{6}}x-1)\]is [SSC (CGL) 2013] |
A) \[0\]
B) \[1\]
C) \[-\,\,1\]
D) \[2\]
Correct Answer: A
Solution :
| Given, \[\cos x+{{\cos }^{2}}x=1\] |
| \[\Rightarrow \] \[cox=1-{{\cos }^{2}}x\] |
| \[\Rightarrow \] \[\cos x={{\sin }^{2}}x\] (i) |
| Now, again \[\cos x+{{\cos }^{2}}x=1\] |
| On cubing both side, we get |
| \[{{(\cos x+{{\cos }^{2}}x)}^{3}}={{(1)}^{3}}\] |
| \[\Rightarrow \]\[{{\cos }^{3}}x+{{({{\cos }^{2}}x)}^{3}}+3{{\cos }^{2}}x{{\cos }^{2}}x\] |
| \[+\,\,3\cos x{{\cos }^{4}}x=1\] |
| \[\Rightarrow \]\[{{\cos }^{3}}x+{{\cos }^{6}}x+3{{\cos }^{4}}x+3{{\cos }^{5}}x=1\] |
| \[\Rightarrow \]\[{{\sin }^{6}}x+{{\sin }^{12}}x+3{{\sin }^{8}}x+3{{\sin }^{10}}x=1\] |
| [from Eq. (i)] |
| \[\therefore \]\[{{\sin }^{12}}x+3{{\sin }^{10}}x+3{{\sin }^{8}}x+{{\sin }^{6}}x-1=0\] |
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