Is prove-cot(θ)cos(θ)-sin(θ)=-cot(θ)sec(θ) ?

The answer to whether prove-cot(θ)cos(θ)-sin(θ)=-cot(θ)sec(θ) is True

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prove-cot(θ)cos(θ)-sin(θ)=-cot(θ)sec(θ)

Solution

prove

- cot (θ) cos (θ) - sin (θ) = - cot (θ) sec (θ)

Solution

True

Solution steps

- cot (θ) cos (θ) - sin (θ) = - cot (θ) sec (θ)

Manipulating left side

- cot (θ) cos (θ) - sin (θ)

Express with sin, cos

- sin (θ) - cos (θ) cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (θ) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- sin (θ) sin (θ) - cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= - sin^{2} (θ) - cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{- 1}{sin ( θ )}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{sin ( θ )}

= - \frac{1}{sin ( θ )}

Manipulating right side

- cot (θ) sec (θ)

Express with sin, cos

- cot (θ) sec (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= - \frac{1}{sin ( θ )}

= - \frac{1}{sin ( θ )}

= - \frac{1}{sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is prove-cot(θ)cos(θ)-sin(θ)=-cot(θ)sec(θ) ?
The answer to whether prove-cot(θ)cos(θ)-sin(θ)=-cot(θ)sec(θ) is True