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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Multiply by

\frac{1 + sin ( A )}{1 + sin ( A )}

= \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{\frac{( 1 - s i n ( A ) ) ( 1 + s i n ( A ) )}{1 + s i n ( A )}}

Apply Difference of Two Squares Formula:

A^{2} - y^{2} = (A + y) (A - y)

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

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

Use the Pythagorean identity:

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

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

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

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

\frac{cos ( A )}{\frac{c o s ^{2} ( A )}{1 + s i n ( A )}}

Apply the fraction rule:

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

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

Cancel the common factor:

cos (A)

= \frac{1 + sin ( A )}{cos ( A )}

= \frac{cos ( A )}{sin ( A ) + 1} + \frac{sin ( A ) + 1}{cos ( A )}

= \frac{cos ( A )}{sin ( A ) + 1} + \frac{sin ( A ) + 1}{cos ( A )}

Simplify

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

\frac{1 + sin ( A )}{cos ( A )} + \frac{cos ( A )}{1 + sin ( A )}

Least Common Multiplier of

cos (A), 1 + sin (A) : cos (A) (sin (A) + 1)

cos (A), 1 + sin (A)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (A)

1 + sin (A)

= cos (A) (sin (A) + 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 (A) (sin (A) + 1)

For

\frac{1 + sin ( A )}{cos ( A )} :

multiply the denominator and numerator by

sin (A) + 1

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

For

\frac{cos ( A )}{1 + sin ( A )} :

multiply the denominator and numerator by

cos (A)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

(1 + sin (A))^{2} : 1 + 2 sin (A) + sin^{2} (A)

Apply Perfect Square Formula:

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

a = 1, b = sin (A)

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

Simplify

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

1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

= 1 + 2 sin (A) + sin^{2} (A)

= 1 + 2 sin (A) + sin^{2} (A)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 2 sin (A) + 1 + 1

Add the numbers:

1 + 1 = 2

= 2 sin (A) + 2

= 2 sin (A) + 2

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

Factor

2 sin (A) + 2 : 2 (sin (A) + 1)

2 sin (A) + 2

Rewrite as

= 2 sin (A) + 2 \cdot 1

Factor out common term

2

= 2 (sin (A) + 1)

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

Cancel the common factor:

sin (A) + 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= 2 sec (A)

2 sec (A)

2 sec (A)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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