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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Apply rule

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

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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