Is 1/(sin(x)cot(x))=sec(x) ?

The answer to whether 1/(sin(x)cot(x))=sec(x) is True

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prove 1/(sin(x)cot(x))=sec(x)

prove

\frac{1}{sin ( x ) cot ( x )} = sec (x)

True

Solution steps

\frac{1}{sin ( x ) cot ( x )} = sec (x)

Manipulating left side

\frac{1}{sin ( x ) cot ( x )}

Express with sin, cos

\frac{1}{cot ( x ) sin ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

\frac{cos ( x )}{sin ( x )} sin (x) : cos (x)

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= cos (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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

cos^{2} (x) = 1 - sin (x) \cdot cos (x) \cdot tan (x)

sin^{2} (x) = \frac{( cos ( 2 x ) - 2 )}{- 4}

cot^{2} (x) \frac{1}{csc ( x )} = csc^{2} (x)

(tan (x)) (1 - cot (x)) = tan (x) - 1

Is 1/(sin(x)cot(x))=sec(x) ?
The answer to whether 1/(sin(x)cot(x))=sec(x) is True