Is cos(x)-sec(x)=-sin(x)tan(x) ?

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

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prove cos(x)-sec(x)=-sin(x)tan(x)

prove

cos (x) - sec (x) = - sin (x) tan (x)

True

Solution steps

cos (x) - sec (x) = - sin (x) tan (x)

Manipulating left side

cos (x) - sec (x)

Express with sin, cos

cos (x) - sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

cos (x) - \frac{1}{cos ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (x) cos (x) - 1

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= cos^{2} (x) - 1

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply the fraction rule:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

- sin (x) tan (x)

- sin (x) tan (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (2 a) - tan (a) = tan (a) sec (2 a)

sin^{2} (x) - 4 cos^{2} (x) = 5 sin^{2} (x) - 4

\frac{cot ( x ) - 1}{cot ( x ) + 1} = \frac{cos ( 2 x )}{1 + sin ( 2 x )}

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

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

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