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

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

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Popular Trigonometry >

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

Solution

prove

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (b) = sin (b)

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

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

Divide fractions:

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

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

Cancel the common factor:

sin (b)

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

Factor out common term

cos (b)

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

Cancel the common factor:

sin (b) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (b)

sec (b)

sec (b)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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