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)

$\frac{1}{x}$

$\frac{1}{\cos 0^{\circ }.\cos 1^{\circ }}+\frac{1}{\cos 1^{\circ }.\cos 2^{\circ }}+\frac{1}{\cos 2^{\circ }.\cos 3^{\circ }}+...+\frac{1}{\cos {44}^{\circ }.\cos {45}^{\circ }}$

$=\frac{1}{x}(\frac{\sin (1^0-0^0)}{\cos 0^{\circ }.\cos 1^{\circ }}+\frac{\sin (2^0-1^0)}{\cos 1^{\circ }.\cos 2^{\circ }}+\frac{\sin (3^0-2^0)}{\cos 2^{\circ }.\cos 3^{\circ }}+...+\frac{\sin ({45}^0-{44}^0)}{\cos {44}^{\circ }.\cos {45}^{\circ }})$

$=\frac{1}{x}(\frac{\sin 1^0\cos 0^0-cos{1}^0\sin 0^0}{\cos 0^{\circ }.\cos 1^{\circ }}+\frac{(\sin 2^0\cos 1^0-\cos 2^0\sin 1^0}{\cos 1^{\circ }.\cos 2^{\circ }}+\frac{\sin 3^0\cos 2^0-cos{3}^0\sin 2^0}{\cos 2^{\circ }.\cos 3^{\circ }}+...+\frac{\sin {45}^0\cos {44}^0-cos{45}^0\sin {44}^0}{\cos {44}^{\circ }.\cos {45}^{\circ }})$..........$\sin (A-B)=\sin A\cos B-\sin B\cos A$

$=\frac{1}{x}(\tan 1^0-\tan 0^0+\tan 2^0-\tan 1^0+$

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