Is (1+cos(θ))/(tan(θ)+sin(θ))=cot(θ) ?

The answer to whether (1+cos(θ))/(tan(θ)+sin(θ))=cot(θ) is True

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prove (1+cos(θ))/(tan(θ)+sin(θ))=cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Apply the fraction rule:

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

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

Factor out common term

sin (θ)

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

Cancel the common factor:

1 + cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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