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

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

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

prove

cot (x) \cdot sec (x) = csc (x)

True

Solution steps

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

Manipulating left side

cot (x) sec (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

2 - csc^{2} (θ) = 1 - cot^{2} (θ)

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

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

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

cot^{4} (x) - csc^{4} (x) = 1 - 2 csc^{2} (x)