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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (- x) - sin (- x) tan (- x)

Use the negative angle identity:

sin (- x) = - sin (x)

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

Use the negative angle identity:

tan (- x) = - tan (x)

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

Use the negative angle identity:

sec (- x) = sec (x)

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

Express with sin, cos

sec (x) - (- sin (x)) (- tan (x))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

- (- a) = a

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (arcsin (\frac{5}{13}))

arcsec (2)

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

- cos (2 π)

arccos (- \frac{1}{2})

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