Is (csc(x)+1)/(cot(x)+cos(x))=sec(x) ?

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

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prove (csc(x)+1)/(cot(x)+cos(x))=sec(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

sin (x)

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

Factor out common term

cos (x)

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

Cancel the common factor:

sin (x) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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