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 right side

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 )}

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

arccsc (x) = sin (x)

cot (θ) = \frac{sin ( θ )}{cos ( θ )}

(1 - tan^{2} (θ))^{2} = sec^{2} (θ) - 4 tan^{2} (θ)

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

4 sin (2 x) = 8 sin (x) cos (x)

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