Is cos(-b)tan(b)csc(b)cot(b)=cot(b) ?

The answer to whether cos(-b)tan(b)csc(b)cot(b)=cot(b) is True

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prove cos(-b)tan(b)csc(b)cot(b)=cot(b)

prove

cos (- b) tan (b) csc (b) cot (b) = cot (b)

True

Solution steps

cos (- b) tan (b) csc (b) cot (b) = cot (b)

Manipulating left side

cos (- b) tan (b) csc (b) cot (b)

Use the negative angle identity:

cos (- x) = cos (x)

= cos (b) cot (b) csc (b) tan (b)

Express with sin, cos

cos (b) cot (b) csc (b) tan (b)

Use the basic trigonometric identity:

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

= cos (b) \frac{cos ( b )}{sin ( b )} csc (b) tan (b)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (b)

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

Cancel the common factor:

sin (b)

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

Multiply:

1 \cdot cos (b) = cos (b)

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

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

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

Use the basic trigonometric identity:

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

= cot (b)

We showed that the two sides could take the same form

\Rightarrow True

Is cos(-b)tan(b)csc(b)cot(b)=cot(b) ?
The answer to whether cos(-b)tan(b)csc(b)cot(b)=cot(b) is True