Is cos(θ)cot(θ)= 1/(sin(θ))-sin(θ) ?

The answer to whether cos(θ)cot(θ)= 1/(sin(θ))-sin(θ) is True

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prove cos(θ)cot(θ)= 1/(sin(θ))-sin(θ)

prove

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

True

Solution steps

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

Manipulating left side

cos (θ) cot (θ)

Express with sin, cos

cos (θ) cot (θ)

Use the basic trigonometric identity:

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

= cos (θ) \frac{cos ( θ )}{sin ( θ )}

Simplify

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{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

\frac{cos ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

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

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

Expand

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

\frac{1 - sin ^{2} ( θ )}{sin ( θ )}

Apply the fraction rule:

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

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

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

Cancel

\frac{sin ^{2} ( θ )}{sin ( θ )} : sin (θ)

\frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= \frac{1}{sin ( θ )} - sin (θ)

= \frac{1}{sin ( θ )} - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

Is cos(θ)cot(θ)= 1/(sin(θ))-sin(θ) ?
The answer to whether cos(θ)cot(θ)= 1/(sin(θ))-sin(θ) is True