Pythagorean Identities - Please Help?

Question

Hello,

I am finding these questions quite challenging:

1) Prove that cosec^2 x(tan^2 x - sin^2 x) ≡ tan^2 x

2) Solve the equation sec^2 x = 4tan x for -π < x < π

3) Solve the equation sec^2 x = 3tan x for -180 < x < 180

Answer 1

• 1).

The trigonometric identity is csc² x (tan² x - sin² x) = tan² x.

Left hand side identity : csc² x (tan² x - sin² x).

Using reciprical identity : csc x = 1/sin x.

= (1/sin² x) (tan² x - sin² x)

= (tan² x/sin² x) - (sin² x/sin² x)

Using quotient identity : tan x = sin x/cos x.

= [ sin² x/(cos² x * sin² x) ] - 1

= (1/cos² x) - 1

= (1 - cos² x)/cos² x

Using pythageran identity : sin² x= 1 - cos² x.

= sin² x/cos² x

= tan² x

= Right hand side identity.

Answer 2

(2). The trigonometric equation sec² x = 4 tan x for - π < x < π.

1 + tan² x = 4 tan x

tan² x - 4 tan x + 1 = 0

Solve the quadratic equation by completing the square.

Separate variables and constants aside by subtracting 1/2 from each side

tan² x - 4 tan x = - 1

Take one half of the coefficient of tan(x) and square it, then add it both sides.

 Here tan(x) coefficient = - 4. so, (half the tan(x))² = (- 4/2)²= 4.

Add 4 to each side.

tan² x - 4 tan x + 4 = - 1 + 4

tan² x - 4 tan x + 4 = - 1 + 4

[tan(x) + 2]²= 3

Take square root both sides.

tan(x) + 2 = ± √3

tan(x) = - 2 ± √3

tan(x) = - 2 + √3 or tan(x) = - 2 - √3

tan(x) = - (2 - √3) or tan(x) = - (2 + √3)

tan(x) = - tan(15^o) or tan(x) = - tan(75^o)

tan(x) = tan(- 15^o) or tan(x) = tan(- 75^o)

tan(x) = tan(- π/12) or tan(x) = tan(- 5π/12)

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

x = nπ + (- π/12) or x = nπ + (- 5π/12)

x = nπ - π/12 or x = nπ - 5π/12.

x = π(n - 1/12) or x = π(n - 5/12).

The solutions outside the interval (- π, π ) are

If n = - 1 then

x = π(- 1 - 1/12) = π[ (- 12 - 1)/12 ] = - 13π/12 = - 195^o or

x = π(- 1 - 5/12) = π[ (- 12 - 5)/12 ] = - 17π/12 = - 255^o.

If n = 2 then

x = π(2 - 1/12) = π[ (24 - 1)/12 ] = 23π/12 = 345^o or

x = π(2 - 5/12) = π[ (24 - 5)/12 ] = 19π/12 = 285^o.

The solutions in the interval (- π, π ) are

If n = 0 then

x = π(0 - 1/12) =  - π/12 = - 15^o or

x = π(0 - 5/12) = - 5π/12 = - 75^o.

If n = 1 then

x = π(1 - 1/12) = π[ (12 - 1)/12 ] = 11π/12 = 165^o or

x = π(1 - 5/12) = π[ (12 - 5)/12 ] = 7π/12 = 105^o.

The solutions in the interval (- π, π ) are - 5π/12, - π/12, 11π/12 and 7π/12.

 

Comments

The equation is [tan(x) - 2]²= 3.

The solutions in the interval (- π, π ) are - 7π/12, - 11π/12, 5π/12 and π/12.

Answer 3

(2). The trigonometric equation sec² x = 4 tan x for - π < x < π.

1 + tan² x = 4 tan x

tan² x - 4 tan x + 1 = 0

Solve the quadratic equation by completing the square.

Separate variables and constants aside by subtracting 1/2 from each side

tan² x - 4 tan x = - 1

Take one half of the coefficient of tan(x) and square it, then add it both sides.

 Here tan(x) coefficient = - 4. so, (half the tan(x))² = (- 4/2)²= 4.

Add 4 to each side.

tan² x - 4 tan x + 4 = - 1 + 4

tan² x - 4 tan x + 4 = - 1 + 4

[tan(x) - 2]²= 3

Take square root both sides.

tan(x) - 2 = ± √3

tan(x) = 2 ± √3

tan(x) = 2 + √3 or tan(x) = 2 - √3

tan(x) = tan(75^o) or tan(x) = -tan(15^o)

tan(x) = tan(5π/12) or tan(x) = tan(π/12)

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

x = nπ + (5π/12) or x = nπ + (π/12)

x = π(n + 5/12) or x = π(n + 1/12).

The solutions outside the interval (- π, π ) are

If n = - 2 then

x = π(- 2 + 5/12) = π[ (- 24 + 5)/12 ] = - 19π/12 = - 285^o or

x = π(- 2 + 1/12) = π[ (- 24 + 1)/12 ] = - 23π/12 = - 345^o.

If n = 1 then

x = π(1 + 5/12) = π[ (12 + 5)/12 ] = 17π/12 = 255^o or

x = π(1 + 1/12) = π[ (12 + 1)/12 ] = 13π/12 = 195^o.

The solutions in the interval (- π, π ) are

If n = - 1 then

x = π(- 1 + 5/12) = π[ (- 12 + 5)/12 ] = - 7π/12 = - 105^o or

x = π(- 1 + 1/12) = π[ (- 12 + 1)/12 ] = - 11π/12 = - 165^o.

If n = 0 then

x = π(0 + 5/12) = 5π/12 = 75^o or

x = π(0 + 1/12) = π/12 = 15^o.

The solutions in the interval (- π, π ) are - 7π/12, - 11π/12, 5π/12 and π/12.

Answer 4

  

• (3). The trigonometric equation is sec² x = 3 tan x for - π < x < π.

1 + tan² x = 3 tan x

tan² x - 3 tan x + 1 = 0

Solve the quadratic equation by completing the square.

Separate variables and constants aside by subtracting 1 from each side.

tan² x - 3 tan x = - 1

Take one half of the coefficient of tan(x) and square it, then add it both sides.

 Here tan(x) coefficient = - 3. so, (half the tan(x))² = (- 3/2)²= 9/4.

Add 9/4 to each side.

tan² x - 3 tan x + 9/4 = - 1 + 9/4

[tan(x) - 3/2]²= 5/4

Take square root both sides.

tan(x) - 3/2 = ± √5/2

tan(x) = [ (3/2) ± (√5/2) ]

tan(x) = (3 + √5)/2 and tan(x) = (3 - √5)/2

tan(x) = tan[ tan^-1(3 + √5)/2 ] and tan(x) = tan[ tan^-1(3 - √5)/2 ]

 

The genaral solution of tan(θ) = tan(α) is θ = 180n + α, where n is an integer.

x = tan^-1 [ (3 + √5)/2 ] + 180n and x = tan^-1 [ (3 - √5)/2 ] + 180n

x = 69.1 + 180n and x =20.9 + 180n                       [ By using calculator ]

 

The solutions outside the interval (- 180, 180 ) are

If n = - 2 then

x = 69.1 + 180(- 2) = 69.1 - 360 = - 290.9  or

x = 20.9 + 180(- 2) = 20.9 - 360 = - 339.1.

If n = 1 then

x = 69.1 + 180(1) = 69.1 + 180 = 249.1  or

x = 20.9 + 180(1) = 20.9 + 180 = 200.9.

 

The solutions in the interval (- 180, 180 ) are

If n = - 1 then

x = 69.1 + 180(- 1) = 69.1 - 180 = - 110.9  or

x = 20.9 + 180(- 1) = 20.9 - 180 = - 159.1.

If n = 0 then

x = 69.1 + 180(0) = 69.1 + 0 = 69.1  or

x = 20.9 + 180(0) = 20.9 + 0 = 20.9.

 

The solutions for - 180 < x < 180 are - 159.1, - 110.9, 20.9 and 69.1.