Is (sin^2(A))/(csc^2(A))=sin^4(A) ?

The answer to whether (sin^2(A))/(csc^2(A))=sin^4(A) is True

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prove (sin^2(A))/(csc^2(A))=sin^4(A)

prove

\frac{sin ^{2} ( A )}{csc ^{2} ( A )} = sin^{4} (A)

True

Solution steps

\frac{sin ^{2} ( A )}{csc ^{2} ( A )} = sin^{4} (A)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{sin ^{2} ( A )}{( \frac{1}{s i n ( A )} ) ^{2}} : sin^{4} (A)

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (A) sin^{2} (A)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{2 + 2} (A)

Add the numbers:

2 + 2 = 4

= sin^{4} (A)

= sin^{4} (A)

= sin^{4} (A)

We showed that the two sides could take the same form

\Rightarrow True

1 - tan^{4} (θ) = 2 sec^{2} (θ) - sec^{4} (θ)

csc (x) sin (2 x) - sec (x) = cos (2 x) sec (x)

(sec (x) + tan (x)) (csc (x) - 1) = cot (x)

cos (\frac{π}{2}) + 1 = sin (\frac{π}{2})

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

Is (sin^2(A))/(csc^2(A))=sin^4(A) ?
The answer to whether (sin^2(A))/(csc^2(A))=sin^4(A) is True