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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

cos (θ) + cot (θ)

Express with sin, cos

cos (θ) + cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

cos (θ) + \frac{cos ( θ )}{sin ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Factor

\frac{cos ( θ ) + cos ( θ ) sin ( θ )}{sin ( θ )} : \frac{cos ( θ ) ( 1 + sin ( θ ) )}{sin ( θ )}

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

Factor

cos (θ) + cos (θ) sin (θ) : cos (θ) (1 + sin (θ))

cos (θ) + cos (θ) sin (θ)

Factor out common term

cos (θ)

= cos (θ) (1 + sin (θ))

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{( 1 + \frac{1}{c s c ( θ )} ) cos ( θ )}{\frac{1}{c s c ( θ )}}

Use the basic trigonometric identity:

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

\frac{( 1 + \frac{1}{c s c ( θ )} ) \frac{1}{s e c ( θ )}}{\frac{1}{c s c ( θ )}}

Simplify

\frac{( 1 + \frac{1}{c s c ( θ )} ) \frac{1}{s e c ( θ )}}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{( 1 + \frac{1}{c s c ( θ )} ) \frac{1}{s e c ( θ )} csc ( θ )}{1}

Join

1 + \frac{1}{csc ( θ )} : \frac{csc ( θ ) + 1}{csc ( θ )}

1 + \frac{1}{csc ( θ )}

Convert element to fraction:

1 = \frac{1 csc ( θ )}{csc ( θ )}

= \frac{1 \cdot csc ( θ )}{csc ( θ )} + \frac{1}{csc ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot csc ( θ ) + 1}{csc ( θ )}

Multiply:

1 \cdot csc (θ) = csc (θ)

= \frac{csc ( θ ) + 1}{csc ( θ )}

= \frac{\frac{c s c ( θ ) + 1}{c s c ( θ )} \cdot \frac{1}{s e c ( θ )} csc ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{csc ( θ ) + 1}{csc ( θ )} \cdot \frac{1}{sec ( θ )} csc (θ)

Multiply fractions:

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

= \frac{( csc ( θ ) + 1 ) \cdot 1 \cdot csc ( θ )}{csc ( θ ) sec ( θ )}

Cancel the common factor:

csc (θ)

= \frac{( csc ( θ ) + 1 ) \cdot 1}{sec ( θ )}

(csc (θ) + 1) \cdot 1 = csc (θ) + 1

(csc (θ) + 1) \cdot 1

Multiply:

(csc (θ) + 1) \cdot 1 = (csc (θ) + 1)

= (csc (θ) + 1)

Remove parentheses:

(a) = a

= csc (θ) + 1

= \frac{csc ( θ ) + 1}{sec ( θ )}

\frac{csc ( θ ) + 1}{sec ( θ )}

\frac{csc ( θ ) + 1}{sec ( θ )}

= \frac{1 + csc ( θ )}{sec ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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