Is cos(θ)=cot(θ)sin^2(θ)csc(θ) ?

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

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prove cos(θ)=cot(θ)sin^2(θ)csc(θ)

prove

cos (θ) = cot (θ) sin^{2} (θ) csc (θ)

True

Solution steps

cos (θ) = cot (θ) sin^{2} (θ) csc (θ)

Manipulating right side

cot (θ) sin^{2} (θ) csc (θ)

Express with sin, cos

cot (θ) csc (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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)

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

Cancel the common factor:

sin^{2} (θ)

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)

\frac{sin ^{2} ( 2 x )}{4} = sin^{2} (x) cos^{2} (x)

20 cos^{2} (x) - cos (x) - 1 = 0

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