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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= 1

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

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

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e (sec (u) - tan (u)) (csc (u) + 1) = cot (u)

p ro v e sin (x) = cos (x - \frac{π}{2})

p ro v e \frac{1}{1 + cos ( x )} = csc^{2} (x) - csc (x) cot (x)

p ro v e \frac{1}{1 - sin ( x )} - \frac{1}{1 + sin ( x )} = \frac{2 tan ( x )}{cos ( x )}

p ro v e cos (t) cot (t) = \frac{1 - sin ^{2} ( t )}{sin ( t )}

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