Is (tan(θ)+sec(θ))(1-sin(θ))=cos(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

(tan (θ) + sec (θ)) (1 - sin (θ))

Express with sin, cos

(1 - sin (θ)) (sec (θ) + tan (θ))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

cot (π - x) = - cot (x)

sin (θ) cos (θ) + sin^{3} (θ) sec (θ) = tan (θ)

tan (A) sin (A) = sec (A) - cos (A)

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

sin (x) sec (x) = 1

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