Is sin(θ)cot^2(θ)=cos^2(θ)csc(θ) ?

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

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prove sin(θ)cot^2(θ)=cos^2(θ)csc(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin (θ) cot^{2} (θ)

Express with sin, cos

cot^{2} (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{cos ^{2} ( θ )}{\frac{1}{c s c ( θ )}}

Simplify

\frac{cos ^{2} ( θ )}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{cos ^{2} ( θ ) csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos^{2} (θ) csc (θ)

cos^{2} (θ) csc (θ)

cos^{2} (θ) csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( x ) tan ( x )}{sec ^{2} ( x )} = sin (x)

sin (α) = \frac{1}{csc ( α )}

sin (2 x) = 1 - cos (2 x)

5 sec^{2} (x) + 4 = 9 sec^{2} (x) - 4 tan^{2} (x)

csc (- x) - sin (- x) = cos (x) cot (x)

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