Is csc(θ)=cot(θ)sec(θ) ?

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

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prove csc(θ)=cot(θ)sec(θ)

prove

csc (θ) = cot (θ) sec (θ)

True

Solution steps

csc (θ) = cot (θ) sec (θ)

Manipulating right side

cot (θ) sec (θ)

Express with sin, cos

cot (θ) sec (θ)

Use the basic trigonometric identity:

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

= \frac{cos ( θ )}{sin ( θ )} sec (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

sec (θ) sin (θ) cot (θ) = 1

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

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

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