Question

Let $x=\sin 1^{\circ }$ , then the value of the expression
$\frac{1}{\cos 0^{\circ }.\cos 1^{\circ }}+\frac{1}{\cos 1^{\circ }.\cos 2^{\circ }}+\frac{1}{\cos 2^{\circ }.\cos 3^{\circ }}+....+\frac{1}{\cos {44}^{\circ }.\cos {45}^{\circ }}$ is equal to

• A. $x$
• B. $\frac{1}{x}$
• C. $\frac{\sqrt{2}}{x}$
• D. $\frac{x}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{1}{x}$

Given $x=\sin 1^{\circ }$

$\frac{1}{\cos 0^{\circ }\cos 1^{\circ }}+\frac{1}{\cos 1^{\circ }\cos 2^{\circ }}+\frac{1}{\cos 2^{\circ }\cos 3^{\circ }}+\cdots +\frac{1}{\cos {44}^{\circ }\cos {45}^{\circ }}=\frac{1}{\sin 1^{\circ }}(\sum _{r=0}^{44}\frac{\sin 1^{\circ }}{\cos r^{\circ }\cos (r+1{)}^{\circ }})$

$=\frac{1}{x}(\sum _{r=0}^{44}\frac{\sin \left(\right(r+1{)}^{\circ }-r^{\circ })}{\cos (r+1{)}^{\circ }\cos r^{\circ }})$

$=\frac{1}{x}\sum _{r=0}^{44}\frac{\sin (r+1{)}^{\circ }\cos r^{\circ }-\cos (r+1{)}^{\circ }\sin r^{\circ }}{\cos (r+1{)}^{\circ }\cos r^{\circ }}$

$=\frac{1}{x}\sum _{r=0}^{44}\left(\tan \right(r+1{)}^{\circ }-\tan r^{\circ })$

$=\frac{1}{x}(\tan 1^{\circ }-\tan 0^{\circ }+\tan 2^{\circ }-\tan 1^{\circ }+\cdots +\tan {45}^{\circ }-\tan {44}^{\circ })$

$=\frac{1}{x}(\tan {45}^{\circ }-\tan 0^{\circ })$

$=\frac{1}{x}$