simplify 1/cos^2x-1/cot^2x

Question

simplify using trig identities

Answer 1

/*** Trig in a Nutshell

Graph a general point P(x,y) in Quadrant I (works in any Quadrant; just more intuitive in I).

Draw a segment from A(0,0) to P(x,y).

Draw a segment from P(x,y) to B(x,0).

Label the segment lengths: AB = x, BP = y, AP = r

By the Pythagorean Theorem r = √(x^2+y^2), where r is defined as positive.

Label the angle BAP as θ.

Since P(x,y) := P( r cos(θ), r sin(θ) ), where ":=" means "is defined as", then:

sin(θ) = y/r
cos(θ) = x/r
tan(θ) := sin(θ)/cos(θ) = y/x
cot(θ) := 1/tan(θ) = x/y
sec(θ) := 1/cos(θ) = r/x
csc(θ) := 1/sin(θ) = r/y

***/

 

Let's use this nutshell to solve the given problem.

 

1/cos^2(θ) - 1/cot^2(θ) =

 

1/(x/r)^2 - 1/(x/y)^2 =

 

(r^2 - y^2)/x^2 =

 

((x^2 + y^2) - y^2)/x^2 =

 

x^2/x^2 = 1

Answer 2

1/cos²x  -  1/cot²x = 1/cos²x - sin²x/cos²x    (cotx = cosx/sinx)

 

=> (1-sin²x)/cos²x = cos²x/cos²x = 1     ( sin²x + cos²x = 1)

 

:)