Is sec(x)cot(x)=csc(x) ?

The answer to whether sec(x)cot(x)=csc(x) is True

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prove sec(x)cot(x)=csc(x)

prove

sec (x) cot (x) = csc (x)

True

Solution steps

sec (x) cot (x) = csc (x)

Manipulating left side

sec (x) cot (x)

Express with sin, cos

cot (x) sec (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (2 x) = \frac{2}{cot ( x ) - tan ( x )}

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

(csc (x) - cot (x)) (sec (x) + 1) = tan (x)

(csc^{2} (x) - 1) sec (x) = cos (x) csc^{2} (x)

(1 + cot^{2} (x)) tan^{2} (x) = sec^{2} (x)