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

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

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

prove

cot (x) + 1 = csc (x) (cos (x) + sin (x))

True

Solution steps

cot (x) + 1 = csc (x) (cos (x) + sin (x))

Manipulating left side

cot (x) + 1

Express with sin, cos

1 + cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

Use the basic trigonometric identity:

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

= (cos (x) + sin (x)) csc (x)

= csc (x) (cos (x) + sin (x))

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{sin ^{2} ( x )}{cos ( x ) - 1} = - 1 - cos (x)

\frac{7 cot ( x )}{sec ( x )} = 7 csc (x) - 7 sin (x)

sin (θ + \frac{π}{3}) = sin (θ) + sin (\frac{π}{3})

(4 sin (θ) - 4 cos (θ))^{2} = 16 - 16 sin (2 θ)

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