Is (tan(θ)cot(θ))/(csc(θ))=sin(θ) ?

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

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prove (tan(θ)cot(θ))/(csc(θ))=sin(θ)

prove

\frac{tan ( θ ) cot ( θ )}{csc ( θ )} = sin (θ)

True

Solution steps

\frac{tan ( θ ) cot ( θ )}{csc ( θ )} = sin (θ)

Manipulating left side

\frac{tan ( θ ) cot ( θ )}{csc ( θ )}

Express with sin, cos

\frac{cot ( θ ) tan ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

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

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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