Is (sec(u)-tan(u))(csc(u)+1)=cot(u) ?

The answer to whether (sec(u)-tan(u))(csc(u)+1)=cot(u) is True

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Popular Trigonometry >

prove (sec(u)-tan(u))(csc(u)+1)=cot(u)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(sec (u) - tan (u)) (csc (u) + 1)

Express with sin, cos

(1 + csc (u)) (sec (u) - tan (u))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

= (1 + \frac{1}{sin ( u )}) (\frac{1}{cos ( u )} - \frac{sin ( u )}{cos ( u )})

Simplify

(1 + \frac{1}{sin ( u )}) (\frac{1}{cos ( u )} - \frac{sin ( u )}{cos ( u )}) : \frac{( sin ( u ) + 1 ) ( 1 - sin ( u ) )}{sin ( u ) cos ( u )}

(1 + \frac{1}{sin ( u )}) (\frac{1}{cos ( u )} - \frac{sin ( u )}{cos ( u )})

Join

1 + \frac{1}{sin ( u )} : \frac{sin ( u ) + 1}{sin ( u )}

1 + \frac{1}{sin ( u )}

Convert element to fraction:

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

= \frac{1 \cdot sin ( u )}{sin ( u )} + \frac{1}{sin ( u )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ( u ) + 1}{sin ( u )}

Multiply:

1 \cdot sin (u) = sin (u)

= \frac{sin ( u ) + 1}{sin ( u )}

= \frac{sin ( u ) + 1}{sin ( u )} (\frac{1}{cos ( u )} - \frac{sin ( u )}{cos ( u )})

Combine the fractions

\frac{1}{cos ( u )} - \frac{sin ( u )}{cos ( u )} : \frac{1 - sin ( u )}{cos ( u )}

Apply rule

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

= \frac{1 - sin ( u )}{cos ( u )}

= \frac{sin ( u ) + 1}{sin ( u )} (\frac{- sin ( u ) + 1}{cos ( u )})

Remove parentheses:

(a) = a

= \frac{sin ( u ) + 1}{sin ( u )} \cdot \frac{1 - sin ( u )}{cos ( u )}

Multiply fractions:

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

= \frac{( sin ( u ) + 1 ) ( 1 - sin ( u ) )}{sin ( u ) cos ( u )}

= \frac{( sin ( u ) + 1 ) ( 1 - sin ( u ) )}{sin ( u ) cos ( u )}

= \frac{( 1 + sin ( u ) ) ( 1 - sin ( u ) )}{cos ( u ) sin ( u )}

Rewrite using trig identities

\frac{( 1 + sin ( u ) ) ( 1 - sin ( u ) )}{cos ( u ) sin ( u )}

Expand

(1 + sin (u)) (1 - sin (u)) : 1 - sin^{2} (u)

(1 + sin (u)) (1 - sin (u))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (u)

= 1^{2} - sin^{2} (u)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (u)

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

Use the Pythagorean identity:

1 = cos^{2} (u) + sin^{2} (u)

1 - sin^{2} (u) = cos^{2} (u)

= \frac{cos ^{2} ( u )}{cos ( u ) sin ( u )}

Cancel the common factor:

cos (u)

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

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

Use the basic trigonometric identity:

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

= cot (u)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (sec(u)-tan(u))(csc(u)+1)=cot(u) ?
The answer to whether (sec(u)-tan(u))(csc(u)+1)=cot(u) is True