What is the general solution for sec^2(θ)+csc^2(θ)=1 ?

The general solution for sec^2(θ)+csc^2(θ)=1 is No Solution for θ\in\mathbb{R}

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Popular >

sec^2(θ)+csc^2(θ)=1

Solution

sec^{2} (θ) + csc^{2} (θ) = 1

Solution

No Solution for θ \in R

Solution steps

sec^{2} (θ) + csc^{2} (θ) = 1

Subtract

1

from both sides

sec^{2} (θ) + csc^{2} (θ) - 1 = 0

Express with sin, cos

- 1 + csc^{2} (θ) + sec^{2} (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= - 1 + (\frac{1}{sin ( θ )})^{2} + sec^{2} (θ)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - 1 + (\frac{1}{sin ( θ )})^{2} + (\frac{1}{cos ( θ )})^{2}

Simplify

- 1 + (\frac{1}{sin ( θ )})^{2} + (\frac{1}{cos ( θ )})^{2} : \frac{- sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

- 1 + (\frac{1}{sin ( θ )})^{2} + (\frac{1}{cos ( θ )})^{2}

(\frac{1}{sin ( θ )})^{2} = \frac{1}{sin ^{2} ( θ )}

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= - 1 + \frac{1}{sin ^{2} ( θ )} + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

1 = \frac{1}{1}

= - \frac{1}{1} + \frac{1}{sin ^{2} ( θ )} + \frac{1}{cos ^{2} ( θ )}

Least Common Multiplier of

1, sin^{2} (θ), cos^{2} (θ) : sin^{2} (θ) cos^{2} (θ)

1, sin^{2} (θ), cos^{2} (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= sin^{2} (θ) cos^{2} (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin^{2} (θ) cos^{2} (θ)

For

\frac{1}{1} :

multiply the denominator and numerator by

sin^{2} (θ) cos^{2} (θ)

\frac{1}{1} = \frac{1 \cdot sin ^{2} ( θ ) cos ^{2} ( θ )}{1 \cdot sin ^{2} ( θ ) cos ^{2} ( θ )} = \frac{sin ^{2} ( θ ) cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

For

\frac{1}{sin ^{2} ( θ )} :

multiply the denominator and numerator by

cos^{2} (θ)

\frac{1}{sin ^{2} ( θ )} = \frac{1 \cdot cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} = \frac{cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

For

\frac{1}{cos ^{2} ( θ )} :

multiply the denominator and numerator by

sin^{2} (θ)

\frac{1}{cos ^{2} ( θ )} = \frac{1 \cdot sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} = \frac{sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

= - \frac{sin ^{2} ( θ ) cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{cos ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

= \frac{- sin ^{2} ( θ ) cos ^{2} ( θ ) + cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ ) cos ^{2} ( θ )}

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ ) - cos ^{2} ( θ ) sin ^{2} ( θ )}{cos ^{2} ( θ ) sin ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos^{2} (θ) + sin^{2} (θ) - cos^{2} (θ) sin^{2} (θ) = 0

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ) - cos^{2} (θ) sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= - cos^{2} (θ) sin^{2} (θ) + 1

1 - cos^{2} (θ) sin^{2} (θ) = 0

Factor

1 - cos^{2} (θ) sin^{2} (θ) : (1 + sin (θ) cos (θ)) (1 - sin (θ) cos (θ))

1 - cos^{2} (θ) sin^{2} (θ)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

sin^{2} (θ) cos^{2} (θ) = (sin (θ) cos (θ))^{2}

= 1 - (sin (θ) cos (θ))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

1 - (sin (θ) cos (θ))^{2} = (sin (θ) cos (θ) + 1) (- sin (θ) cos (θ) + 1)

= (sin (θ) cos (θ) + 1) (- sin (θ) cos (θ) + 1)

(1 + sin (θ) cos (θ)) (1 - sin (θ) cos (θ)) = 0

Solving each part separately

1 + sin (θ) cos (θ) = 0 or 1 - sin (θ) cos (θ) = 0

1 + sin (θ) cos (θ) = 0 :

No Solution

1 + sin (θ) cos (θ) = 0

Rewrite using trig identities

1 + sin (θ) cos (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= 1 + \frac{sin ( 2 θ )}{2}

1 + \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

1 + \frac{sin ( 2 θ )}{2} = 0

Subtract

1

from both sides

1 + \frac{sin ( 2 θ )}{2} - 1 = 0 - 1

Simplify

\frac{sin ( 2 θ )}{2} = - 1

\frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

\frac{2 sin ( 2 θ )}{2} = 2 (- 1)

Simplify

sin (2 θ) = - 2

sin (2 θ) = - 2

- 1 \leq sin (x) \leq 1

No Solution

1 - sin (θ) cos (θ) = 0 :

No Solution

1 - sin (θ) cos (θ) = 0

Rewrite using trig identities

1 - sin (θ) cos (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= 1 - \frac{sin ( 2 θ )}{2}

1 - \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

1 - \frac{sin ( 2 θ )}{2} = 0

Subtract

1

from both sides

1 - \frac{sin ( 2 θ )}{2} - 1 = 0 - 1

Simplify

- \frac{sin ( 2 θ )}{2} = - 1

- \frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

- \frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

2 (- \frac{sin ( 2 θ )}{2}) = 2 (- 1)

Simplify

- sin (2 θ) = - 2

- sin (2 θ) = - 2

Divide both sides by

- 1

- sin (2 θ) = - 2

Divide both sides by

- 1

\frac{- sin ( 2 θ )}{- 1} = \frac{- 2}{- 1}

Simplify

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

Popular Examples

cos(t)-sin(t)=0

cos (t) - sin (t) = 0

sin(x)-cos(x)= 1/3

sin (x) - cos (x) = \frac{1}{3}

tan^2(x)=sec(x)-1,0<= x<= 2pi

tan^{2} (x) = sec (x) - 1, 0 \leq x \leq 2 π

sin^2(θ)=2sin^2(θ/2)

sin^{2} (θ) = 2 sin^{2} (\frac{θ}{2})

cos(x)=sqrt(cos(x))

cos (x) = cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for sec^2(θ)+csc^2(θ)=1 ?
The general solution for sec^2(θ)+csc^2(θ)=1 is No Solution for θ\in\mathbb{R}