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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

sec^{2} (x) - sec (x) tan (x)

Express with sin, cos

sec^{2} (x) - sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Apply rule

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

= \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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Frequently Asked Questions (FAQ)

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