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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

cos (β) + \frac{cos ( β )}{sin ( β )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

1 + \frac{1}{sin ( β )}

Convert element to fraction:

1 = \frac{1 sin ( β )}{sin ( β )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (β) = sin (β)

= \frac{sin ( β ) + 1}{sin ( β )}

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

Divide fractions:

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

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

Cancel the common factor:

sin (β)

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

Factor out common term

cos (β)

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

Cancel the common factor:

sin (β) + 1

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( β )}}

Simplify

\frac{1}{\frac{1}{s e c ( β )}}

Apply the fraction rule:

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

= \frac{sec ( β )}{1}

Apply rule

\frac{a}{1} = a

= sec (β)

sec (β)

sec (β)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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