Is cot(θ)sin(θ)cos(θ)=1-sin^2(θ) ?

The answer to whether cot(θ)sin(θ)cos(θ)=1-sin^2(θ) is True

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prove cot(θ)sin(θ)cos(θ)=1-sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

cot (θ) sin (θ) cos (θ)

Express with sin, cos

cos (θ) cot (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (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)

= cos^{2} (θ)

= cos^{2} (θ)

Rewrite using trig identities

cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 1 - sin^{2} (θ)

= 1 - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

Is cot(θ)sin(θ)cos(θ)=1-sin^2(θ) ?
The answer to whether cot(θ)sin(θ)cos(θ)=1-sin^2(θ) is True