Is csc^2(θ)*(sin^2(θ))=1 ?

The answer to whether csc^2(θ)*(sin^2(θ))=1 is True

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prove csc^2(θ)*(sin^2(θ))=1

prove

csc^{2} (θ) \cdot (sin^{2} (θ)) = 1

True

Solution steps

csc^{2} (θ) sin^{2} (θ) = 1

Manipulating left side

csc^{2} (θ) sin^{2} (θ)

Express with sin, cos

csc^{2} (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

tan (b) cos (b) + cot (b) cos (b) = csc (b)

1 + cot^{2} (z) = csc^{2} (z)

csc (x) - 1 = \frac{2 sin ( x )}{sin ( x )}

2 sin^{2} (θ) - cos (2 θ) = 1

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