Is 1/(tan(θ)csc(θ))=cos(θ) ?

The answer to whether 1/(tan(θ)csc(θ))=cos(θ) is True

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prove 1/(tan(θ)csc(θ))=cos(θ)

prove

\frac{1}{tan ( θ ) csc ( θ )} = cos (θ)

True

Solution steps

\frac{1}{tan ( θ ) csc ( θ )} = cos (θ)

Manipulating left side

\frac{1}{tan ( θ ) csc ( θ )}

Express with sin, cos

\frac{1}{csc ( θ ) tan ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Cancel the common factor:

sin (θ)

= \frac{1}{cos ( θ )}

= \frac{1}{\frac{1}{c o s ( θ )}}

Apply the fraction rule:

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

= \frac{cos ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

\frac{1 + sec ( x )}{tan ( x ) + sin ( x )} = csc (x)

sec (θ) + tan (θ) = \frac{1}{sec ( θ ) - tan ( θ )}

Is 1/(tan(θ)csc(θ))=cos(θ) ?
The answer to whether 1/(tan(θ)csc(θ))=cos(θ) is True