Is csc(θ)-cot(θ)*cos(θ)=sin(θ) ?

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

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prove csc(θ)-cot(θ)*cos(θ)=sin(θ)

prove

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

True

Solution steps

csc (θ) - cot (θ) cos (θ) = sin (θ)

Manipulating left side

csc (θ) - cot (θ) cos (θ)

Express with sin, cos

csc (θ) - cos (θ) cot (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (θ) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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