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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

(1 + sin (x)) (sec (x) - tan (x))

Express with sin, cos

(1 + sin (x)) (sec (x) - tan (x))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

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

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

tan (\frac{π}{6}) = \frac{1}{3}

cos (4 a) = 1 - 8 sin^{2} (a) cos^{2} (a)

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