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

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

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Popular Trigonometry >

prove 1/(1+sin(x))=(sec(x)-tan(x))sec(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

(sec (x) - tan (x)) sec (x)

Express with sin, cos

(sec (x) - tan (x)) sec (x)

Use the basic trigonometric identity:

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

= (\frac{1}{cos ( x )} - tan (x)) \frac{1}{cos ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

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

Apply rule

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

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

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

Remove parentheses:

(a) = a

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

Multiply fractions:

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

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

(1 - sin (x)) \cdot 1 = 1 - sin (x)

(1 - sin (x)) \cdot 1

Multiply:

(1 - sin (x)) \cdot 1 = (1 - sin (x))

= (1 - sin (x))

Remove parentheses:

(a) = a

= 1 - sin (x)

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - sin^{2} (x)

Factor out common term

- 1

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

Factor

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

sin^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Cancel

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

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

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

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

Refine

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

Cancel the common factor:

1 - sin (x)

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

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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