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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( x )} + tan ( x )}{1 + csc ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

1 + sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (a) = \frac{sec ( a )}{csc ( a )}

sec (x) tan (x) cos (x) cot (x) = 1

1 - sec (x) cos^{3} (x) = sin^{2} (x)

sin (2 x) (1 - cos (2 x)) = 4 sin^{3} (x) cos (x)

cot (θ) - 2 cot (2 θ) = tan (θ)

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