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

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

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

prove

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

True

Solution steps

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

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{π}{3}) = cos (\frac{π}{6})

cos^{2} (t) = \frac{( 1 + cos ( 2 t ) )}{2}

cos (- \frac{π}{6}) = sin (\frac{2 π}{3})

sin (x + π) - sin (x) + 1 = 0

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

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