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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{sin ( θ ) cos ( θ )}

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \frac{1}{cos ( θ )}

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

Join

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

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

Least Common Multiplier of

sin (θ) cos (θ), cos (θ) : sin (θ) cos (θ)

sin (θ) cos (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ) cos (θ)

cos (θ)

= sin (θ) cos (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (θ) cos (θ)

For

\frac{1}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (θ)

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

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

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