Is (1+sec(β))/(tan(β)+sin(β))=csc(β) ?

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

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prove (1+sec(β))/(tan(β)+sin(β))=csc(β)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1 + \frac{1}{c o s ( β )}}{sin ( β ) + tan ( β )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

sin (β) + \frac{sin ( β )}{cos ( β )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

1 + \frac{1}{cos ( β )}

Convert element to fraction:

1 = \frac{1 cos ( β )}{cos ( β )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (β) = cos (β)

= \frac{cos ( β ) + 1}{cos ( β )}

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

Divide fractions:

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

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

Cancel the common factor:

cos (β)

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

Factor out common term

sin (β)

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

Cancel the common factor:

cos (β) + 1

= \frac{1}{sin ( β )}

= \frac{1}{sin ( β )}

= \frac{1}{sin ( β )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( β )}}

Simplify

\frac{1}{\frac{1}{c s c ( β )}}

Apply the fraction rule:

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

= \frac{csc ( β )}{1}

Apply rule

\frac{a}{1} = a

= csc (β)

csc (β)

csc (β)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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