Is csc^2(b)-csc^2(b)cos^2(b)=1 ?

The answer to whether csc^2(b)-csc^2(b)cos^2(b)=1 is True

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prove csc^2(b)-csc^2(b)cos^2(b)=1

prove

csc^{2} (b) - csc^{2} (b) cos^{2} (b) = 1

True

Solution steps

csc^{2} (b) - csc^{2} (b) cos^{2} (b) = 1

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\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 )}

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

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

(\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 )} cos^{2} (b)

Multiply fractions:

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

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

Multiply:

1 \cdot cos^{2} (b) = cos^{2} (b)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

cos (\frac{5 θ}{2}) cos (\frac{θ}{2}) = \frac{1}{4} cos^{2} (5 θ)

sin (3 v) = 3 sin (v) - 4 sin^{3} (v)

2 sin (x + \frac{π}{4}) = sin (x) + cos (x)

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

(3 cos (θ))^{2} + (3 sin (θ))^{2} = 9

Is csc^2(b)-csc^2(b)cos^2(b)=1 ?
The answer to whether csc^2(b)-csc^2(b)cos^2(b)=1 is True