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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (x) sec (x)

Express with sin, cos

cos (x) sec (x)

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

= 1

= 1

Manipulating right side

tan (x) cot (x)

Express with sin, cos

cot (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

\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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

csc (\frac{π}{2} - x) cot (x) = csc (x)

sec (\frac{3 π}{2} - x) = - csc (x)

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