Question

The integer part of the number $\sum _{k=0}^{44}\frac{1}{\cos k^{\circ }\cos (k+1{)}^{\circ }}$ is

• A. $50$
• B. $52$
• C. $57$
• D. $59$

Answer 1

The correct option is C $57$

$S=\sum _{k=0}^{44}\frac{1}{\cos k^{\circ }\cos (k+1{)}^{\circ }}$
Mutliplying and dividing the expression with $\sin 1^{\circ },$
$S=\frac{1}{\sin 1^{\circ }}\sum _{k=0}^{44}\frac{\sin 1^{\circ }}{\cos k^{\circ }\cos (k+1{)}^{\circ }}\Rightarrow S=\frac{1}{\sin 1^{\circ }}\sum _{k=0}^{44}\frac{\sin \left(\right(k+1)-k{)}^{\circ }}{\cos k^{\circ }\cos (k+1{)}^{\circ }}\Rightarrow S=\frac{1}{\sin 1^{\circ }}\sum _{k=0}^{44}\left(\tan \right(k+1{)}^{\circ }-\tan k^{\circ })\Rightarrow S=\frac{1}{\sin 1^{\circ }}(\tan {45}^{\circ }-\tan 0^{\circ })\Rightarrow S=\frac{1}{\sin 1^{\circ }}=\frac{1}{\sin \frac{\pi }{180}}=57.29$

Hence the integer part of the number is $57.$