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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Combine the fractions

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

Apply rule

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

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

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

Remove parentheses:

(a) = a

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

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

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

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\frac{1 - sin ^{4} ( x )}{cos ^{2} ( x )} = 2 - cos^{2} (x)

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

prove tan(x)cos(x)=sin(x)prove

tan (x) cos (x) = sin (x)

Frequently Asked Questions (FAQ)

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