Is cos(θ)-sec(θ)=-sin(θ)tan(θ) ?

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

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prove cos(θ)-sec(θ)=-sin(θ)tan(θ)

prove

cos (θ) - sec (θ) = - sin (θ) tan (θ)

True

Solution steps

cos (θ) - sec (θ) = - sin (θ) tan (θ)

Manipulating left side

cos (θ) - sec (θ)

Express with sin, cos

cos (θ) - sec (θ)

Use the basic trigonometric identity:

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

= cos (θ) - \frac{1}{cos ( θ )}

Simplify

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

cos (θ) - \frac{1}{cos ( θ )}

Convert element to fraction:

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

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

cos (θ) cos (θ) - 1 = cos^{2} (θ) - 1

cos (θ) cos (θ) - 1

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)

= cos^{2} (θ) - 1

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{- sin ^{2} ( θ )}{cos ( θ )}

Apply the fraction rule:

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

= - \frac{sin ^{2} ( θ )}{cos ( θ )}

= - \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

- sin (θ) tan (θ)

- sin (θ) tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan (a) + cot (a) = sec (a) csc (a)

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

\frac{cot ( x ) sec ( x )}{csc ( x )} = 1

1 + cot^{2} (θ) = csc^{2} (θ)

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

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