Is (tan(x)+sin(x))/(1+sec(x))=sin(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (tan(x)+sin(x))/(1+sec(x))=sin(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

1 + \frac{1}{cos ( x )} : \frac{cos ( x ) + 1}{cos ( x )}

1 + \frac{1}{cos ( x )}

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (x)

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

Factor out common term

sin (x)

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

Cancel the common factor:

cos (x) + 1

= sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( 3 a ) cos ( 2 a )}{sin ( 3 a ) sin ( 2 a )} = \frac{cot ( 2 a )}{tan ( 3 a )}

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

(sin (θ) - 1) (tan (θ) + sec (θ)) = - cos (θ)

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

sin (b) cos^{2} (b) - sin (b) = - sin^{3} (b)

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