Is (csc(β))/(cot(β))=sec(β) ?

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

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prove (csc(β))/(cot(β))=sec(β)

prove

\frac{csc ( β )}{cot ( β )} = sec (β)

True

Solution steps

\frac{csc ( β )}{cot ( β )} = sec (β)

Manipulating left side

\frac{csc ( β )}{cot ( β )}

Express with sin, cos

\frac{csc ( β )}{cot ( β )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( β )}}{cot ( β )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin (β)

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

= \frac{1}{cos ( β )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( β )}}

Simplify

\frac{1}{\frac{1}{s e c ( β )}}

Apply the fraction rule:

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

= \frac{sec ( β )}{1}

Apply rule

\frac{a}{1} = a

= sec (β)

sec (β)

sec (β)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin^{2} (\frac{θ}{2}) = \frac{csc ( θ ) - cot ( θ )}{2 csc ( θ )}

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

sin^{2} (x) - 2 cos^{2} (x) = 3 sin^{2} (x) - 2

Is (csc(β))/(cot(β))=sec(β) ?
The answer to whether (csc(β))/(cot(β))=sec(β) is True