Is csc(θ)=(sec(θ))/(tan(θ)) ?

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

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prove csc(θ)=(sec(θ))/(tan(θ))

prove

csc (θ) = \frac{sec ( θ )}{tan ( θ )}

True

Solution steps

csc (θ) = \frac{sec ( θ )}{tan ( θ )}

Manipulating right side

\frac{sec ( θ )}{tan ( θ )}

Express with sin, cos

\frac{sec ( θ )}{tan ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( θ )}}{tan ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( θ )}}{\frac{s i n ( θ )}{c o s ( θ )}}

Simplify

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

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

Divide fractions:

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

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

Refine

= \frac{cos ( θ )}{cos ( θ ) sin ( 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

tan^{2} (B) - sin^{2} (B) = tan^{2} (B) sin^{2} (B)

\frac{cos ( 2 x ) + 1}{sin ( 2 x )} = cot (x)

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

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

\frac{cot ( x )}{tan ( x ) sec ( x )} = csc (x) - 1

Is csc(θ)=(sec(θ))/(tan(θ)) ?
The answer to whether csc(θ)=(sec(θ))/(tan(θ)) is True