Is ((cos(a-b)))/(sin(a)sin(b))=cot(a)cot(b)+1 ?

The answer to whether ((cos(a-b)))/(sin(a)sin(b))=cot(a)cot(b)+1 is True

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prove ((cos(a-b)))/(sin(a)sin(b))=cot(a)cot(b)+1

prove

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

True

Solution steps

\frac{cos ( a - b )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

Manipulating left side

\frac{cos ( a - b )}{sin ( a ) sin ( b )}

Rewrite using trig identities

\frac{cos ( a - b )}{sin ( a ) sin ( b )}

Use the Angle Difference identity:

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

= \frac{cos ( a ) cos ( b ) + sin ( a ) sin ( b )}{sin ( a ) sin ( b )}

= \frac{cos ( a ) cos ( b ) + sin ( a ) sin ( b )}{sin ( a ) sin ( b )}

Manipulating right side

cot (a) cot (b) + 1

Express with sin, cos

1 + cot (a) cot (b)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

1 + \frac{cos ( a )}{sin ( a )} \cdot \frac{cos ( b )}{sin ( b )} : \frac{sin ( a ) sin ( b ) + cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

1 + \frac{cos ( a )}{sin ( a )} \cdot \frac{cos ( b )}{sin ( b )}

Multiply

\frac{cos ( a )}{sin ( a )} \cdot \frac{cos ( b )}{sin ( b )} : \frac{cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

\frac{cos ( a )}{sin ( a )} \cdot \frac{cos ( b )}{sin ( b )}

Multiply fractions:

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

= \frac{cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

= 1 + \frac{cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (a) = sin (a)

= \frac{sin ( a ) sin ( b ) + cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

= \frac{sin ( a ) sin ( b ) + cos ( a ) cos ( b )}{sin ( a ) sin ( b )}

= \frac{cos ( a ) cos ( b ) + sin ( a ) sin ( b )}{sin ( a ) sin ( b )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin^{2} (x) = 3 cos^{2} (x)

(sin (a) + cos (a))^{2} = 1 + sin (2 a)

2 cos (x) (sin (3 x) - sin (x)) = sin (4 x)

\frac{1 - sin ( θ )}{sec ( θ )} = \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

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