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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

tan (\frac{π}{2} - x) sec (x) = csc (x)

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

sec (u) cos (u) - sin^{2} (u) = cos^{2} (u)

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

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