Is ((sec(θ)-tan(θ))^2+1)/(csc(θ)(sec(θ)-tan(θ)))=2tan(θ) ?

The answer to whether ((sec(θ)-tan(θ))^2+1)/(csc(θ)(sec(θ)-tan(θ)))=2tan(θ) is True

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prove ((sec(θ)-tan(θ))^2+1)/(csc(θ)(sec(θ)-tan(θ)))=2tan(θ)

Solution

prove

\frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{csc ( θ ) ( sec ( θ ) - tan ( θ ) )} = 2 tan (θ)

Solution

True

Solution steps

\frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{csc ( θ ) ( sec ( θ ) - tan ( θ ) )} = 2 tan (θ)

Manipulating left side

\frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{csc ( θ ) ( sec ( θ ) - tan ( θ ) )}

Express with sin, cos

\frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{( sec ( θ ) - tan ( θ ) ) csc ( θ )}

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( θ )} - tan ( θ ) ) ^{2} + 1}{( \frac{1}{c o s ( θ )} - tan ( θ ) ) csc ( θ )}

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) csc ( θ )}

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) \frac{1}{s i n ( θ )}}

Simplify

\frac{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) \frac{1}{s i n ( θ )}} : \frac{sin ( θ ) ( cos ^{2} ( θ ) + ( - sin ( θ ) + 1 ) ^{2} )}{cos ( θ ) ( 1 - sin ( θ ) )}

\frac{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) \frac{1}{s i n ( 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{( \frac{1}{c o s ( θ )} - \frac{s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{( \frac{- s i n ( θ ) + 1}{c o s ( θ )} ) \frac{1}{s i n ( 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{( \frac{- s i n ( θ ) + 1}{c o s ( θ )} ) ^{2} + 1}{( \frac{- s i n ( θ ) + 1}{c o s ( θ )} ) \frac{1}{s i n ( θ )}}

Remove parentheses:

(a) = a

= \frac{( \frac{1 - s i n ( θ )}{c o s ( θ )} ) ^{2} + 1}{\frac{1 - s i n ( θ )}{c o s ( θ )} \cdot \frac{1}{s i n ( θ )}}

Multiply

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

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

Multiply fractions:

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

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

(1 - sin (θ)) \cdot 1 = 1 - sin (θ)

(1 - sin (θ)) \cdot 1

Multiply:

(1 - sin (θ)) \cdot 1 = (1 - sin (θ))

= (1 - sin (θ))

Remove parentheses:

(a) = a

= 1 - sin (θ)

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

= \frac{( \frac{- s i n ( θ ) + 1}{c o s ( θ )} ) ^{2} + 1}{\frac{1 - s i n ( θ )}{c o s ( θ ) s i n ( θ )}}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{\frac{( - s i n ( θ ) + 1 ) ^{2}}{c o s ^{2} ( θ )} + 1}{\frac{1 - s i n ( θ )}{c o s ( θ ) s i n ( θ )}}

Apply the fraction rule:

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

= \frac{( \frac{( 1 - s i n ( θ ) ) ^{2}}{c o s ^{2} ( θ )} + 1 ) cos ( θ ) sin ( θ )}{1 - sin ( θ )}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

= \frac{\frac{c o s ^{2} ( θ ) + ( - s i n ( θ ) + 1 ) ^{2}}{c o s ^{2} ( θ )} cos ( θ ) sin ( θ )}{1 - sin ( θ )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

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

= \frac{\frac{s i n ( θ ) ( c o s ^{2} ( θ ) + ( - s i n ( θ ) + 1 ) ^{2} )}{c o s ( θ )}}{1 - sin ( θ )}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

(1 - sin (θ))^{2} + 1 - sin^{2} (θ)

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

Apply Perfect Square Formula:

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

a = 1, b = sin (θ)

= 1^{2} - 2 \cdot 1 \cdot sin (θ) + sin^{2} (θ)

Simplify

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

1^{2} - 2 \cdot 1 \cdot sin (θ) + sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot sin (θ) + sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 1 - 2 sin (θ) + sin^{2} (θ)

= 1 - 2 sin (θ) + sin^{2} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

sin^{2} (θ) - sin^{2} (θ) = 0

= - 2 sin (θ) + 1 + 1

Add the numbers:

1 + 1 = 2

= - 2 sin (θ) + 2

= - 2 sin (θ) + 2

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

Factor

- 2 sin (θ) + 2 : 2 (- sin (θ) + 1)

- 2 sin (θ) + 2

Rewrite as

= - 2 sin (θ) + 2 \cdot 1

Factor out common term

2

= 2 (- sin (θ) + 1)

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

Cancel the common factor:

- sin (θ) + 1

= \frac{2 sin ( θ )}{cos ( θ )}

= \frac{2 sin ( θ )}{cos ( θ )}

= \frac{2 sin ( θ )}{cos ( θ )}

Rewrite using trig identities

= 2 \cdot \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

2 tan (θ)

2 tan (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))^2+1)/(csc(θ)(sec(θ)-tan(θ)))=2tan(θ) ?
The answer to whether ((sec(θ)-tan(θ))^2+1)/(csc(θ)(sec(θ)-tan(θ)))=2tan(θ) is True