Is sin^2(θ)(csc^2(θ)-1)=cos^2(θ) ?

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

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prove sin^2(θ)(csc^2(θ)-1)=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin^{2} (θ) (csc^{2} (θ) - 1)

Express with sin, cos

(- 1 + csc^{2} (θ)) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

(- 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 )}

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

Join

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

- 1 + \frac{1}{sin ^{2} ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= - - 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

cot (α) sec (α) = csc (α)

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

\frac{1}{sec ( a ) + tan ( a )} = sec (a) - tan (a)

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

sin^{3} (θ) + sin (θ) cos^{2} (θ) = sin (θ)

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