Is sin(θ)cos(θ)csc^2(θ)=cot(θ) ?

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

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prove sin(θ)cos(θ)csc^2(θ)=cot(θ)

prove

sin (θ) cos (θ) csc^{2} (θ) = cot (θ)

True

Solution steps

sin (θ) cos (θ) csc^{2} (θ) = cot (θ)

Manipulating left side

sin (θ) cos (θ) csc^{2} (θ)

Express with sin, cos

cos (θ) csc^{2} (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

Multiply fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

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

Cancel the common factor:

sin (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

csc (2 θ) = \frac{1 + tan ^{2} ( θ )}{2 tan ( θ )}

tan^{2} (x) (1 + cos (2 x)) = 1 - cos (2 x)

\frac{1 + tan ( θ )}{sin ( θ ) + cos ( θ )} = sec (θ)

\frac{cos ( x ) sec ( x )}{cot ( x )} = tan (x)

8 cos^{4} (θ - 8) sin^{4} (θ) = 8 cos (2 θ)

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