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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

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

Manipulating right side

1 + cot (x) tan (y)

Express with sin, cos

1 + cot (x) tan (y)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( y )}{cos ( y )} : \frac{cos ( x ) sin ( y )}{sin ( x ) cos ( y )}

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( y )}{cos ( y )}

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

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 ( y ) + cos ( x ) sin ( y )}{sin ( x ) cos ( y )}

Multiply:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (2 x) = \frac{1}{2} - \frac{1}{2} cos (4 x)

\frac{1}{1 + tan ( x ) tan ( 2 x )} = cos (2 x)

sin (\frac{a}{2}) = \frac{1}{2} sin (a)

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

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