Is 1+cos(x)=(csc(x)+cot(x))/(csc(x)) ?

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

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prove 1+cos(x)=(csc(x)+cot(x))/(csc(x))

prove

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

True

Solution steps

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

Manipulating right side

\frac{csc ( x ) + cot ( x )}{csc ( x )}

Express with sin, cos

\frac{cot ( x ) + csc ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )} + csc ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Divide fractions:

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

= \frac{( cos ( x ) + 1 ) sin ( x )}{sin ( x ) \cdot 1}

Refine

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

Cancel the common factor:

sin (x)

= cos (x) + 1

= cos (x) + 1

= 1 + cos (x)

We showed that the two sides could take the same form

\Rightarrow True

(tan (x)) (sin (x)) + (cos (x)) = sec (x)

cos (k (x) x) = cos (- k (x) x)

tan^{2} (x) = sin^{2} (x) + sin^{2} (x) tan^{2} (x)

- sin (x) sec (x) = - tan (x)

cos (4 x) - cos (3 x) cos (x) = sin (2 x) - 4 cos (2 x) sin (2 x)

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