Is (tan^2(A))/(tan^2(A)+1)=sin^2(A) ?

The answer to whether (tan^2(A))/(tan^2(A)+1)=sin^2(A) is True

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prove (tan^2(A))/(tan^2(A)+1)=sin^2(A)

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( A )}{tan ^{2} ( A ) + 1}

Express with sin, cos

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

Simplify

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

= \frac{sin ^{2} ( A )}{\frac{c o s ^{2} ( A ) + s i n ^{2} ( A )}{c o s ^{2} ( A )} cos ^{2} ( A )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (A)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (A)

= sin^{2} (A)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ^{4} ( α ) - 1}{tan ^{2} ( α )} = sec^{2} (α) + 1

\frac{1 + sec ( x )}{csc ( x )} - tan (x) = sin (x)

2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

cos (8 x) = cos^{2} (4 x) - sin^{2} (4 x)

tan (\frac{π}{2} - x) cot (x) = csc^{2} (x) - 1

Is (tan^2(A))/(tan^2(A)+1)=sin^2(A) ?
The answer to whether (tan^2(A))/(tan^2(A)+1)=sin^2(A) is True