A) 1
B) 0
C) 2
D) -1
Correct Answer: D
Solution :
We have \[\cos {{1}^{o}}+\cos {{2}^{o}}+\cos {{3}^{o}}+...+\cos {{180}^{o}}\] \[\Rightarrow \] \[\cos {{1}^{o}}+\cos {{2}^{o}}+\cos {{3}^{o}}+...\cos {{89}^{o}}\] \[+\cos {{90}^{o}}+\cos {{91}^{o}}+\cos {{92}^{o}}+\cos {{93}^{o}}+....\] \[+\cos {{179}^{o}}+\cos {{180}^{o}}\] Now using \[[\cos \,({{180}^{o}}-\theta )=-\cos \theta ]\] we have \[\cos {{1}^{o}}+\cos {{2}^{o}}+\cos {{3}^{o}}+...\] \[\cos {{89}^{o}}+0+\cos \,({{180}^{o}}-{{89}^{o}})\] \[+\cos \,({{180}^{o}}-{{88}^{o}})+...+\cos \,({{180}^{o}}-{{1}^{o}})-1\] [as \[\cos 90=0\] and \[cos\,\,180=-1\]] \[=\cos {{1}^{o}}+\cos {{2}^{o}}+\cos {{3}^{o}}+...\cos {{89}^{o}}-\cos {{89}^{o}}\] \[-\cos {{88}^{o}}-...-\cos {{1}^{o}}-1\] \[=-1\]
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