Is csc(θ)=(tan(θ)cot(θ))/(sin(θ)) ?

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

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prove csc(θ)=(tan(θ)cot(θ))/(sin(θ))

prove

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

True

Solution steps

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

Manipulating right side

\frac{tan ( θ ) cot ( θ )}{sin ( θ )}

Express with sin, cos

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= \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

sin (x) cos (x) - 2 tan (x) = tan (x) cos (2 x)

cot^{2} (x) - cos^{2} (x) = cot^{2} (x) cot (x)

sin^{2} (3 2^{\circ}) + cos^{2} (x) = 1

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

cos (θ) = - \frac{3}{4}

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