Is (1-csc^2(A))/(csc^2(A))=-cos^2(A) ?

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

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prove (1-csc^2(A))/(csc^2(A))=-cos^2(A)

prove

\frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )} = - cos^{2} (A)

True

Solution steps

\frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )} = - cos^{2} (A)

Manipulating left side

\frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )}

Express with sin, cos

\frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )}

Use the basic trigonometric identity:

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

= \frac{1 - ( \frac{1}{s i n ( A )} ) ^{2}}{( \frac{1}{s i n ( A )} ) ^{2}}

Simplify

\frac{1 - ( \frac{1}{s i n ( A )} ) ^{2}}{( \frac{1}{s i n ( A )} ) ^{2}} : sin^{2} (A) - 1

\frac{1 - ( \frac{1}{s i n ( A )} ) ^{2}}{( \frac{1}{s i n ( A )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1 - ( \frac{1}{s i n ( A )} ) ^{2}}{\frac{1}{s i n ^{2} ( A )}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1 - \frac{1}{s i n ^{2} ( A )}}{\frac{1}{s i n ^{2} ( A )}}

Apply the fraction rule:

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

= \frac{( 1 - \frac{1}{s i n ^{2} ( A )} ) sin ^{2} ( A )}{1}

Join

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

1 - \frac{1}{sin ^{2} ( A )}

Convert element to fraction:

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

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

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} ( A ) - 1}{sin ^{2} ( A )}

Multiply:

1 \cdot sin^{2} (A) = sin^{2} (A)

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

= \frac{\frac{s i n ^{2} ( A ) - 1}{s i n ^{2} ( A )} sin ^{2} ( A )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (A)

= sin^{2} (A) - 1

= sin^{2} (A) - 1

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

Rewrite using trig identities

- 1 + sin^{2} (A)

Use the Pythagorean identity:

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

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

= - cos^{2} (A)

= - cos^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

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