Is (cos(θ))/(sin^2(θ))=csc(θ)cot(θ) ?

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

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prove (cos(θ))/(sin^2(θ))=csc(θ)cot(θ)

prove

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

True

Solution steps

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

Manipulating right side

csc (θ) cot (θ)

Express with sin, cos

cot (θ) csc (θ)

Use the basic trigonometric identity:

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

= \frac{cos ( θ )}{sin ( θ )} csc (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

cos (θ) \cdot 1 = cos (θ)

= \frac{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{cos ( θ )}{sin ^{2} ( θ )}

= \frac{cos ( θ )}{sin ^{2} ( θ )}

= \frac{cos ( θ )}{sin ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{2} \frac{tan ( b )}{1 - cos ^{2} ( b )} = csc (2 b)

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

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

\frac{4 + 2 sin ( θ )}{1 - sin ( θ )} = 3

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

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