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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

sin (x)

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

Factor out common term

cos (x)

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

Cancel the common factor:

sin (x) + 1

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (x) = sin^{2} (x)

1 = 1 + cot (θ)

\frac{sin ^{2} ( A )}{csc ^{2} ( A )} = sin^{4} (A)

1 - tan^{4} (θ) = 2 sec^{2} (θ) - sec^{4} (θ)

csc (x) sin (2 x) - sec (x) = cos (2 x) sec (x)

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