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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( x )} ) ^{2} - tan ^{2} ( x )}{1 - sin ( x )}

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( x )} ) ^{2} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 - sin ( x )}

Simplify

\frac{( \frac{1}{c o s ( x )} ) ^{2} - ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 - sin ( x )} : \frac{sin ( x ) + 1}{cos ^{2} ( x )}

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

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

Apply exponent rule:

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

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

= \frac{\frac{1}{c o s ^{2} ( x )} - \frac{s i n ^{2} ( x )}{c o s ^{2} ( x )}}{1 - sin ( x )}

Combine the fractions

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

Apply rule

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

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

= \frac{\frac{1 - s i n ^{2} ( x )}{c o s ^{2} ( x )}}{1 - sin ( x )}

Apply the fraction rule:

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

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

Factor

1 - sin (x) : - (sin (x) - 1)

1 - sin (x)

Factor out common term

- 1

= - (sin (x) - 1)

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

Cancel the common factor:

sin (x) - 1

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

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

= \frac{1 + sin ( x )}{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+sin(x))/(cos^2(x)) ?
The answer to whether 1/(1-sin(x))=(1+sin(x))/(cos^2(x)) is True