Is csc(θ)=(cot(θ))/(cos(θ)) ?

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

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prove csc(θ)=(cot(θ))/(cos(θ))

prove

csc (θ) = \frac{cot ( θ )}{cos ( θ )}

True

Solution steps

csc (θ) = \frac{cot ( θ )}{cos ( θ )}

Manipulating right side

\frac{cot ( θ )}{cos ( θ )}

Express with sin, cos

\frac{cot ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( θ )}{s i n ( θ )}}{cos ( θ )}

Simplify

\frac{\frac{c o s ( θ )}{s i n ( θ )}}{cos ( θ )} : \frac{1}{sin ( θ )}

\frac{\frac{c o s ( θ )}{s i n ( θ )}}{cos ( θ )}

Apply the fraction rule:

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

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

Cancel the common factor:

cos (θ)

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{1 + cos ( s )} = csc^{2} (s) - csc (s) cot (s)

(5 sin (θ) + 5 cos (θ))^{2} = 25 + 25 sin (2 θ)

arccos (x) = \frac{1}{cos ( x )}

\frac{csc ^{2} ( x ) - 1}{cos ( x )} = cot (x) csc (x)

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

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