Is (csc(α)+cot(α))(sec(α)-1)=tan(α) ?

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(csc (α) + cot (α)) (sec (α) - 1)

Express with sin, cos

(- 1 + sec (α)) (cot (α) + csc (α))

Use the basic trigonometric identity:

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

= (- 1 + \frac{1}{cos ( α )}) (cot (α) + csc (α))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

- 1 + \frac{1}{cos ( α )} : \frac{- cos ( α ) + 1}{cos ( α )}

- 1 + \frac{1}{cos ( α )}

Convert element to fraction:

1 = \frac{1 cos ( α )}{cos ( α )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (α) = cos (α)

= \frac{- cos ( α ) + 1}{cos ( α )}

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

Combine the fractions

\frac{cos ( α )}{sin ( α )} + \frac{1}{sin ( α )} : \frac{cos ( α ) + 1}{sin ( α )}

Apply rule

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

= \frac{cos ( α ) + 1}{sin ( α )}

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

Remove parentheses:

(a) = a

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (α)) (1 - cos (α))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (α)

= 1^{2} - cos^{2} (α)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (α)

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

Use the Pythagorean identity:

1 = cos^{2} (α) + sin^{2} (α)

1 - cos^{2} (α) = sin^{2} (α)

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

Cancel the common factor:

sin (α)

= \frac{sin ( α )}{cos ( α )}

= \frac{sin ( α )}{cos ( α )}

Use the basic trigonometric identity:

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

= tan (α)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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