Is 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x) ?

The answer to whether 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x) is True

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Popular Trigonometry >

prove 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x)

Solution

prove

\frac{1}{1 + sin ( x )} + \frac{1}{1 - sin ( x )} = 2 + 2 tan^{2} (x)

Solution

True

Solution steps

\frac{1}{1 + sin ( x )} + \frac{1}{1 - sin ( x )} = 2 + 2 tan^{2} (x)

Manipulating left side

\frac{1}{1 + sin ( x )} + \frac{1}{1 - sin ( x )}

Simplify

\frac{1}{1 + sin ( x )} + \frac{1}{1 - sin ( x )} : \frac{2}{( sin ( x ) + 1 ) ( - sin ( x ) + 1 )}

\frac{1}{1 + sin ( x )} + \frac{1}{1 - sin ( x )}

Least Common Multiplier of

1 + sin (x), 1 - sin (x) : (sin (x) + 1) (- sin (x) + 1)

1 + sin (x), 1 - sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (x)

1 - sin (x)

= (sin (x) + 1) (- sin (x) + 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

(sin (x) + 1) (- sin (x) + 1)

For

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

multiply the denominator and numerator by

- sin (x) + 1

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

For

\frac{1}{1 - sin ( x )} :

multiply the denominator and numerator by

sin (x) + 1

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

= \frac{- sin ( x ) + 1}{( sin ( x ) + 1 ) ( - sin ( x ) + 1 )} + \frac{sin ( x ) + 1}{( sin ( x ) + 1 ) ( - sin ( x ) + 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( x ) + 1 + sin ( x ) + 1}{( sin ( x ) + 1 ) ( - sin ( x ) + 1 )}

- sin (x) + 1 + sin (x) + 1 = 2

- sin (x) + 1 + sin (x) + 1

Group like terms

= - sin (x) + sin (x) + 1 + 1

Add similar elements:

- sin (x) + sin (x) = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

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

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

Rewrite using trig identities

\frac{2}{( 1 + sin ( x ) ) ( 1 - sin ( x ) )}

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

= \frac{2}{cos ^{2} ( x )}

= \frac{2}{cos ^{2} ( x )}

Manipulating right side

2 + 2 tan^{2} (x)

Factor

2 + 2 tan^{2} (x) : 2 (1 + tan^{2} (x))

2 + 2 tan^{2} (x)

Rewrite as

= 2 \cdot 1 + 2 tan^{2} (x)

Factor out common term

2

= 2 (1 + tan^{2} (x))

= (1 + tan^{2} (x)) \cdot 2

Express with sin, cos

(1 + tan^{2} (x)) \cdot 2

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

= 2 \cdot \frac{cos ^{2} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}

Multiply fractions:

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

= \frac{( cos ^{2} ( x ) + sin ^{2} ( x ) ) \cdot 2}{cos ^{2} ( x )}

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

= \frac{( cos ^{2} ( x ) + sin ^{2} ( x ) ) \cdot 2}{cos ^{2} ( x )}

Rewrite using trig identities

\frac{( cos ^{2} ( x ) + sin ^{2} ( x ) ) \cdot 2}{cos ^{2} ( x )}

Use the Pythagorean identity:

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

= \frac{1 \cdot 2}{cos ^{2} ( x )}

Simplify

= \frac{2}{cos ^{2} ( x )}

= \frac{2}{cos ^{2} ( x )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x) ?
The answer to whether 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x) is True