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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply exponent rule:

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

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

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

Multiply

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

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

Cancel the common factor:

cos (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= cos (x) sin (x)

= cos (x) sin (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e 2 csc (2 x) = \frac{1}{sin ( x ) cos ( x )}

p ro v e (1 + csc (θ)) (1 - sin (θ)) = csc (θ) - sin (θ)

p ro v e \frac{1 - tan ( x )}{1 - cot ( x )} = - tan (x)

p ro v e \frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

p ro v e 1 + cot^{2} (- x) = csc^{2} (x)

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