Is (sin(x)-cos(x))/(sin(x))=1-cot(x) ?

The answer to whether (sin(x)-cos(x))/(sin(x))=1-cot(x) is True

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prove (sin(x)-cos(x))/(sin(x))=1-cot(x)

prove

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

True

Solution steps

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

Manipulating right side

1 - cot (x)

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 )}

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{( sec ( θ ) - 1 )}{( 1 - cos ( θ ) )} = sec (θ)

cos (0) tan (0) = sin (0)

cos (\frac{π}{6}) = \frac{4}{3} cos^{3} (\frac{π}{6})

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

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

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