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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{cos ( x )}{sec ( x ) + tan ( x )}

Express with sin, cos

\frac{cos ( x )}{sec ( x ) + tan ( x )}

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{\frac{1}{c o s ( x )} + tan ( x )}

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{\frac{1}{c o s ( x )} + \frac{s i n ( x )}{c o s ( x )}}

Simplify

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

\frac{cos ( x )}{\frac{1}{c o s ( x )} + \frac{s i n ( x )}{c o s ( 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{cos ( x )}{\frac{1 + s i n ( x )}{c o s ( x )}}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Cancel the common factor:

sin (x) + 1

= - (sin (x) - 1)

Negate

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

= - sin (x) + 1

= - sin (x) + 1

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