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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

1 + \frac{1}{cos ( B )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (B) = cos (B)

= \frac{cos ( B ) + 1}{cos ( B )}

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

Divide fractions:

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

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

Cancel the common factor:

cos (B)

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

Factor out common term

sin (B)

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

Cancel the common factor:

cos (B) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( B )}}

Simplify

\frac{1}{\frac{1}{c s c ( B )}}

Apply the fraction rule:

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

= \frac{csc ( B )}{1}

Apply rule

\frac{a}{1} = a

= csc (B)

csc (B)

csc (B)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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