What is the general solution for sin(θ)cos(θ)=csc(θ)sec(θ) ?

The general solution for sin(θ)cos(θ)=csc(θ)sec(θ) is No Solution for θ\in\mathbb{R}

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sin(θ)cos(θ)=csc(θ)sec(θ)

Solution

sin (θ) cos (θ) = csc (θ) sec (θ)

Solution

No Solution for θ \in R

Solution steps

sin (θ) cos (θ) = csc (θ) sec (θ)

Subtract

csc (θ) sec (θ)

from both sides

sin (θ) cos (θ) - csc (θ) sec (θ) = 0

Express with sin, cos

cos (θ) sin (θ) - csc (θ) sec (θ)

Use the basic trigonometric identity:

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

= cos (θ) sin (θ) - \frac{1}{sin ( θ )} sec (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

cos (θ) sin (θ) - \frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )}

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

\frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

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

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{sin ( θ ) cos ( θ )}

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

Convert element to fraction:

cos (θ) sin (θ) = \frac{cos ( θ ) sin ( θ ) sin ( θ ) cos ( θ )}{sin ( θ ) cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (θ) sin (θ) sin (θ) cos (θ) - 1

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

cos (θ) sin (θ) sin (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= sin (θ) sin (θ) cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin (θ) sin (θ) cos^{2} (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ) cos^{2} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ) cos^{2} (θ)

= sin^{2} (θ) cos^{2} (θ) - 1

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

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

\frac{- 1 + cos ^{2} ( θ ) sin ^{2} ( θ )}{cos ( θ ) sin ( θ )} = 0

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

- 1 + cos^{2} (θ) sin^{2} (θ) = 0

Factor

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

- 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)

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

Solving each part separately

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

sin (θ) cos (θ) + 1 = 0 :

No Solution

sin (θ) cos (θ) + 1 = 0

Rewrite using trig identities

sin (θ) cos (θ) + 1

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

sin (θ) cos (θ) - 1 = 0 :

No Solution

sin (θ) cos (θ) - 1 = 0

Rewrite using trig identities

sin (θ) cos (θ) - 1

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

Add

1

to 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} = 1 \cdot 2

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)cos(θ)=csc(θ)sec(θ) ?
The general solution for sin(θ)cos(θ)=csc(θ)sec(θ) is No Solution for θ\in\mathbb{R}