Is sin^2(β)csc^2(β)-sin^2(β)=cos^2(β) ?

The answer to whether sin^2(β)csc^2(β)-sin^2(β)=cos^2(β) is True

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prove sin^2(β)csc^2(β)-sin^2(β)=cos^2(β)

prove

sin^{2} (β) csc^{2} (β) - sin^{2} (β) = cos^{2} (β)

True

Solution steps

sin^{2} (β) csc^{2} (β) - sin^{2} (β) = cos^{2} (β)

Manipulating left side

sin^{2} (β) csc^{2} (β) - sin^{2} (β)

Express with sin, cos

- sin^{2} (β) + csc^{2} (β) sin^{2} (β)

Use the basic trigonometric identity:

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

= - sin^{2} (β) + (\frac{1}{sin ( β )})^{2} sin^{2} (β)

Simplify

- sin^{2} (β) + (\frac{1}{sin ( β )})^{2} sin^{2} (β) : - sin^{2} (β) + 1

- sin^{2} (β) + (\frac{1}{sin ( β )})^{2} sin^{2} (β)

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

(\frac{1}{sin ( β )})^{2} sin^{2} (β)

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

(\frac{1}{sin ( β )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( β )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( β )}

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (β)

= 1

= - sin^{2} (β) + 1

= - sin^{2} (β) + 1

= 1 - sin^{2} (β)

Rewrite using trig identities

1 - sin^{2} (β)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

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

= cos^{2} (β)

= cos^{2} (β)

We showed that the two sides could take the same form

\Rightarrow True

(sin (θ) + 1) (sin (θ) - 1) = - cos^{2} (θ)

cos (x) = sin (x)

csc (x) - cot (x) = tan (\frac{x}{2})

cot^{2} (a) - cos^{2} (a) = cot^{2} (a) cos^{2} (a)

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

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