When we graph the equation x$^{2}$ + y$^{2}$ = 1 in a rectangular coordinates system we will get a circle. This graph is called unit circle with center as origin(0,0) and radius as 1 unit.Trigonometric Functions are the functions for which domain is a set of angles and range is a real numbers. Circular functions are the functions whose domain are the set of numbers that correspond to the measures of angles of with respect to the trigonometric functions. The ranges of these circular functions are the set of numbers that are real numbers with respect to the trigonometric functions.Trigonometric functions are called circular-functions because the radian measures of the angles are calculated by the arc length of the circle. The unit circle x$^{2}$ + y$^{2}$ = 1 shown in the above diagram.Let A be the point on the X-axis where circle touches the X-axis. So the coordinate of point A will be (1,0). Let P be any point on the circle whose coordinates be (x,y). The distance of point P is greater than zero (p>0) for the counter clockwise and in clockwise direction P <0. The x-coordinate of the point P will be define as cosine and Y coordinate of the point P will be sine.

sin p = y

cos p = x

∴ tan p =$\frac{sin p }{cos p}$; cos p $\neq $0

cot p =$\frac{cos p }{sin p}$; sin p $\neq $0

sec p =$\frac{1 }{cos p}$; cos p $\neq $0

csc p =$\frac{1 }{sin p}$; sin p $\neq $0

The coordinates of point P are (cos p, sin p) where cos p and sin p exists for all real numbers. Point P is located according to the arc length . Arc is positive if counted counterclockwise and it is negative if counted clockwise. The range of the circular function is [-1,1].

tan q = $\frac{sin q}{cos q}$ = $\frac{1/3}{\sqrt{8}/3}$ = $\frac{\sqrt{2}}{4}$

cot q = $\frac{cos q}{sin q}$ = $\frac{\sqrt{8}/3}{1/3}$ = $\sqrt{8}$

sec q = $\frac{1}{cos q}$ = $\frac{1}{\sqrt{8}/3}$ = $\frac{3}{\sqrt{8}}$

csc q = $\frac{1}{sin q}$ = $\frac{1}{1/3}$ = 3

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