Is sec(θ)cot(θ)-sin(θ)=cos(θ)cot(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (θ) cot (θ) - sin (θ)

Express with sin, cos

- sin (θ) + cot (θ) sec (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{1}{sin ( θ )}

= - sin (θ) + \frac{1}{sin ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

- sin (θ) sin (θ) + 1 = - sin^{2} (θ) + 1

- sin (θ) sin (θ) + 1

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= - sin^{2} (θ) + 1

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

= cos (θ) \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

cos (θ) cot (θ)

cos (θ) cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

9 sin (9 π - x) = 9 sin (x)

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

sec (x) \cdot cot (x) = csc (x)

\frac{tan ^{2} ( x )}{sec ( t )} = sec (t) - cos (t)

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

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