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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

Join

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

1 + \frac{1}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

Divide fractions:

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

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

Cancel the common factor:

sin (θ)

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

Factor out common term

cos (θ)

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

Cancel the common factor:

sin (θ) + 1

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

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

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

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