Is (sec(θ)-tan(θ))(csc(θ)+1)=cot(θ) ?

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(sec (θ) - tan (θ)) (csc (θ) + 1)

Express with sin, cos

(1 + csc (θ)) (sec (θ) - tan (θ))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{1}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

Combine the fractions

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

Apply rule

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

= \frac{1 - sin ( θ )}{cos ( θ )}

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

Remove parentheses:

(a) = a

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

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

Cancel the common factor:

cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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