Is (1-sin(v))/(cos(v))+(cos(v))/(1-sin(v))=2sec(v) ?

The answer to whether (1-sin(v))/(cos(v))+(cos(v))/(1-sin(v))=2sec(v) is True

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prove (1-sin(v))/(cos(v))+(cos(v))/(1-sin(v))=2sec(v)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Least Common Multiplier of

cos (v), 1 - sin (v) : cos (v) (- sin (v) + 1)

cos (v), 1 - sin (v)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (v)

1 - sin (v)

= cos (v) (- sin (v) + 1)

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

cos (v) (- sin (v) + 1)

For

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

multiply the denominator and numerator by

- sin (v) + 1

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

For

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

multiply the denominator and numerator by

cos (v)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

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

Apply Perfect Square Formula:

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

a = 1, b = sin (v)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

sin^{2} (v) - sin^{2} (v) = 0

= - 2 sin (v) + 1 + 1

Add the numbers:

1 + 1 = 2

= - 2 sin (v) + 2

= - 2 sin (v) + 2

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

Factor

- 2 sin (v) + 2 : 2 (- sin (v) + 1)

- 2 sin (v) + 2

Rewrite as

= - 2 sin (v) + 2 \cdot 1

Factor out common term

2

= 2 (- sin (v) + 1)

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

Cancel the common factor:

- sin (v) + 1

= \frac{2}{cos ( v )}

= \frac{2}{cos ( v )}

= \frac{2}{cos ( v )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{2}{\frac{1}{s e c ( v )}}

Simplify

\frac{2}{\frac{1}{s e c ( v )}}

Apply the fraction rule:

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

= \frac{2 sec ( v )}{1}

Apply rule

\frac{a}{1} = a

= 2 sec (v)

2 sec (v)

2 sec (v)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (1-sin(v))/(cos(v))+(cos(v))/(1-sin(v))=2sec(v) ?
The answer to whether (1-sin(v))/(cos(v))+(cos(v))/(1-sin(v))=2sec(v) is True