Is cot(θ)=cos(θ)csc(θ) ?

The answer to whether cot(θ)=cos(θ)csc(θ) is True

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prove cot(θ)=cos(θ)csc(θ)

prove

cot (θ) = cos (θ) csc (θ)

True

Solution steps

cot (θ) = cos (θ) csc (θ)

Manipulating right side

cos (θ) csc (θ)

Express with sin, cos

csc (θ) cos (θ)

Use the basic trigonometric identity:

csc (θ) = \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )} cos (θ)

Simplify

\frac{1}{sin ( θ )} cos (θ) : \frac{cos ( θ )}{sin ( θ )}

\frac{1}{sin ( θ )} cos (θ)

Multiply fractions:

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

= \frac{1 cos ( θ )}{sin ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

\frac{cos ( θ )}{sin ( θ )} = cot (θ)

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

sech^{2} (x) = 1 - tanh^{2} (x)

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

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

\frac{5 cot ( x )}{sec ( x )} = 5 csc (x) - 5 sin (x)

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