Question

Let $x=\sin 1^{\circ},$ then the value of the expression $\dfrac{1}{\cos 0^{\circ} \cdot \cos 1^{\circ}}+\dfrac{1}{\cos 1^{\circ} \cdot \cos 2^{\circ}}+$
$\dfrac{1}{\cos 2^{\circ} \cdot \cos 3^{\circ}}+\ldots+\dfrac{1}{\cos 44^{\circ} \cdot \cos 45^{\circ}}$ is equal to
A.$\mathrm{x}$
B.$\dfrac{1}{x}$
C.$\dfrac{\sqrt{2}}{x}$
D.$\dfrac{x}{\sqrt{2}}$

Answer 1

Hint: There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as $\sin , \cos ,$ tan, $\mathrm{csc},$ sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle $\theta$. Any cosine function can be written as a sine function. $\mathrm{y}=\mathrm{A} \sin (\mathrm{Bx})$ and $\mathrm{y}=\mathrm{A} \cos (\mathrm{Bx}) .$ The number, $\mathrm{A},$ in front of sine or cosine changes the height of the graph. The value $\mathrm{A}$ (in front of sin or cos) affects the amplitude (height).

Complete step-by-step answer:
The Greek letter theta $\theta$ is used in mathematics as a variable usually associated with
a measured angle. For example, the symbol theta appears in the three main trigonometric functions: sine, cosine, and tangent as the input variable. S
Sine and cosine $\sin (\theta)$ and $\cos (\theta)-$ are functions revealing the shape of a right triangle. Looking out from a vertex with angle $\theta, \sin (\theta)$ is the ratio of the opposite side to the hypotenuse, while $\cos (\theta)$ is the ratio of the adjacent side to the hypotenuse.
Given $\mathrm{x}=\sin 1^{\circ}$
$\dfrac{1}{\cos {{0}^{{}^\circ }}\cos {{1}^{{}^\circ }}}+\dfrac{1}{\cos {{1}^{{}^\circ }}\cos {{2}^{{}^\circ }}}+\dfrac{1}{\cos {{2}^{{}^\circ }}\cos {{3}^{{}^\circ }}}+\cdots +\dfrac{1}{\cos {{44}^{{}^\circ }}\cos {{45}^{{}^\circ }}}$
 Or,$\dfrac{1}{\sin 1^{\circ}}\left(\sum_{\mathrm{r}=0}^{44} \dfrac{\sin 1^{\circ}}{\cos \mathrm{r}^{\circ} \cos (\mathrm{r}+1)^{\circ}}\right)$
Now we can write that:
$\dfrac{1}{\mathrm{x}}\left(\sum_{\mathrm{r}=0}^{44} \dfrac{\sin \left((\mathrm{r}+1)^{\circ}-\mathrm{r}^{\circ}\right)}{\cos (\mathrm{r}+1)^{\circ} \cos \mathrm{r}^{\circ}}\right)$
Or, $\dfrac{1}{\mathrm{x}} \sum_{\mathrm{r}=0}^{44} \dfrac{\sin (\mathrm{r}+1)^{\circ} \cos \mathrm{r}^{\circ}-\cos (\mathrm{r}+1)^{\circ} \sin \mathrm{r}^{\circ}}{\cos (\mathrm{r}+1)^{\circ} \cos \mathrm{r}^{\circ}}$
$=\dfrac{1}{\text{x}}\sum\limits_{\text{r}=0}^{44}{\left( \tan {{(\text{r}+1)}^{{}^\circ }}- \right.}\tan {{r}^{{}^\circ }}$
$=\dfrac{1}{\text{x}}\left( \tan {{1}^{{}^\circ }}-\tan {{0}^{{}^\circ }}+ \right.\left. \tan {{2}^{{}^\circ }}-\tan {{1}^{{}^\circ }}+\cdots +\tan {{45}^{{}^\circ }}-\tan {{44}^{{}^\circ }} \right)$
$=\dfrac{1}{\mathrm{x}}\left(\tan 45^{\circ}-\tan 0^{\circ}\right)$
$=\dfrac{1}{\mathrm{x}}$
So, the correct answer is Option B.

Note: The shape of the sine curve is the same for each full rotation of the angle and so the function is called 'periodic'. The period of the function is $360^{\circ}$ or $2 \pi$ radians. We can rotate the point as many times as we like. In mathematical terms we say the 'domain' of the sine function is the set of all real numbers.
The cosine function is a periodic function which is very important in trigonometry. The simplest way to understand the cosine function is to use the unit circle. The $\mathrm{x}$ -coordinate of the point where the other side of the angle intersects the circle is $\cos (\theta),$ and the $y$ -coordinate is $\sin (\theta)$.