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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (x) cot (x) - cot (x) cos (x)

Express with sin, cos

- cos (x) cot (x) + cot (x) sec (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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)

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (x) \cdot cot (x) = csc (x) - sin (x)

cos (x + 27 0^{\circ}) = sin (x)

cos (u) sec (u) - sin^{2} (u) = cos^{2} (u)

cos^{4} (x) = cos^{2} (x) cos^{2} (x)

cot (x) - 2 cot (2 x) = tan (x)

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