Is (csc(θ)-cot(θ))(1+sec(θ))=tan(θ) ?

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(csc (θ) - cot (θ)) (1 + sec (θ))

Express with sin, cos

(- cot (θ) + csc (θ)) (1 + sec (θ))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Apply rule

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

= \frac{- cos ( θ ) + 1}{sin ( θ )}

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

Join

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

1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + cos (θ)) (1 - cos (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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