Is (cot(x)tan(x))/(cos(x))=sec(x) ?

The answer to whether (cot(x)tan(x))/(cos(x))=sec(x) is True

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prove (cot(x)tan(x))/(cos(x))=sec(x)

prove

\frac{cot ( x ) tan ( x )}{cos ( x )} = sec (x)

True

Solution steps

\frac{cot ( x ) tan ( x )}{cos ( x )} = sec (x)

Manipulating left side

\frac{cot ( x ) tan ( x )}{cos ( x )}

Express with sin, cos

\frac{cot ( x ) tan ( x )}{cos ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )} tan ( x )}{cos ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )} \cdot \frac{s i n ( x )}{c o s ( x )}}{cos ( x )}

Simplify

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

\frac{\frac{c o s ( x )}{s i n ( x )} \cdot \frac{s i n ( x )}{c o s ( x )}}{cos ( x )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Cancel the common factor:

sin (x)

= 1

= \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

\frac{sin ( θ )}{cos ( θ )} = tan (θ)

cos (x) tan (x) csc (x) = 1

sin (6 x) tan (3 x) = 2 sin^{2} (3 x)

cos (θ) cot (θ) + sin^{2} (θ) csc (θ) = csc (θ)

sin^{2} (x) sec^{2} (x) + 1 = sec^{2} (x)

Is (cot(x)tan(x))/(cos(x))=sec(x) ?
The answer to whether (cot(x)tan(x))/(cos(x))=sec(x) is True