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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (b) csc^{2} (b) - sin^{2} (b)

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (b)

= 1

= - sin^{2} (b) + 1

= - sin^{2} (b) + 1

= 1 - sin^{2} (b)

Rewrite using trig identities

1 - sin^{2} (b)

Use the Pythagorean identity:

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

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

= cos^{2} (b)

= cos^{2} (b)

We showed that the two sides could take the same form

\Rightarrow True

(tan (θ) + sec (θ)) (1 - sin (θ)) = cos (θ)

cot (π - x) = - cot (x)

sin (θ) cos (θ) + sin^{3} (θ) sec (θ) = tan (θ)

tan (A) sin (A) = sec (A) - cos (A)

\frac{1}{sin ( x )} = sin (x) + cos (x) cot (x)

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