Math Examples of Trigonometric Functions

Values of Trigonometric Functions can be evaluated from right angled triangle. The secant of a function is represented as follows:
Sec x = Hypotenuse (Longest side) / Base of the triangle (shorter side).... equation 1
 
 
We can use above formula to calculate other trigonometric functions. That is, value of sec x is given; rest of the functions values can be calculated. Let us suppose an example of this to understand solving Trigonometric functions.
Example: If sec x equal p plus 1 by 4p, then how to find the value of sec x plus tan x.
Solution: In the given problem we have the value of sec x = (p + 1) / 4p and we have to calculate the value of sec x + tan x.
For this we first have to calculate the value of tan x using the value of sec x. On comparing the value of sec x i.e. (p + 1) / 4p we see that hypotenuse or the longest side of the Right Triangle is equals to (p + 1) and base or one of the two shorter sides is equals to 4p. We can also use direct trigonometric identity:
Sec² x = 1 + tan² x,
((p + 1) / 4p) 2 = 1 + tan² x,
(p² + 1 + 2p) /16p² – 1 =tan² x,
Or tan x = √ (-15p² + 2p + 1) / 4p,
Or we can also use pythagoras theorem to find the value of tan x.
Perpendicular side² = Hypotenuse² – Base² = p² + 1 + 2p - 16p² = -15p² + 2p + 1,
Or Perpendicular side = √ (-15p² + 2p + 1),
Value of tan x = √ (-15p² + 2p + 1) / 4p.