Is (tan^2(θ))/(1+tan^2(θ))=sin^2(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{sin ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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cos^{2} (θ) + cot^{2} (θ) cos^{2} (θ) = cot^{2} (θ)

\frac{sec ^{2} ( a )}{2 - sec ^{2} ( a )} = sec (2 a)

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