Is (tan^2(A))/(sec^2(A))=sin^2(A) ?

The answer to whether (tan^2(A))/(sec^2(A))=sin^2(A) is True

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prove (tan^2(A))/(sec^2(A))=sin^2(A)

prove

\frac{tan ^{2} ( A )}{sec ^{2} ( A )} = sin^{2} (A)

True

Solution steps

\frac{tan ^{2} ( A )}{sec ^{2} ( A )} = sin^{2} (A)

Manipulating left side

\frac{tan ^{2} ( A )}{sec ^{2} ( A )}

Express with sin, cos

\frac{tan ^{2} ( A )}{sec ^{2} ( A )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{( \frac{s i n ( A )}{c o s ( A )} ) ^{2}}{sec ^{2} ( A )}

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

cos^{2} (A)

= sin^{2} (A)

= sin^{2} (A)

= sin^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

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