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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{sin ( x ) + tan ( 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

(sec (x) - tan (x)) (csc (x) + 1) = cot (x)

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

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

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

(1 + cot^{2} (x)) tan (x) = csc (x) sec (x)

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