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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (θ) csc^{2} (θ) - sin^{2} (θ)

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

- sin^{2} (θ) + (\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

= - sin^{2} (θ) + 1

= - sin^{2} (θ) + 1

= 1 - sin^{2} (θ)

Rewrite using trig identities

1 - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (x + \frac{π}{3}) + sin (x - \frac{π}{6}) = 0

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

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

\frac{8}{cos ( x ) + 1} + \frac{8}{cos ( x ) - 1} = - 16 csc (x) cot (x)

sec (x) (sec (x) - cos (x)) = tan^{2} (x)

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