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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Combine the fractions

\frac{1}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )} : \frac{1 + sin ( θ )}{cos ( θ )}

Apply rule

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

= \frac{1 + sin ( θ )}{cos ( θ )}

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

Apply the fraction rule:

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

= \frac{cos ( θ ) cos ( θ )}{1 + sin ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( θ )}{1 + sin ( θ )}

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

1 - sin^{2} (θ)

Factor out common term

- 1

= - (sin^{2} (θ) - 1)

Factor

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

sin^{2} (θ) - 1

Rewrite

1

1^{2}

= sin^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sin (θ) + 1) (sin (θ) - 1)

= - (sin (θ) + 1) (sin (θ) - 1)

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

Cancel the common factor:

sin (θ) + 1

= - (sin (θ) - 1)

Negate

- (sin (θ) - 1) = - sin (θ) + 1

= - sin (θ) + 1

= - sin (θ) + 1

= - sin (θ) + 1

= 1 - sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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