Is 1/(1-sin(B))=sec^2(B)+sec(B)tan(B) ?

The answer to whether 1/(1-sin(B))=sec^2(B)+sec(B)tan(B) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 1/(1-sin(B))=sec^2(B)+sec(B)tan(B)

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (B) + sec (B) tan (B)

Express with sin, cos

sec^{2} (B) + sec (B) tan (B)

Use the basic trigonometric identity:

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

= (\frac{1}{cos ( B )})^{2} + \frac{1}{cos ( B )} tan (B)

Use the basic trigonometric identity:

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

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

Simplify

(\frac{1}{cos ( B )})^{2} + \frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )} : \frac{1 + sin ( B )}{cos ^{2} ( B )}

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

(\frac{1}{cos ( B )})^{2} = \frac{1}{cos ^{2} ( B )}

(\frac{1}{cos ( B )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( B )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( B )}

\frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )} = \frac{sin ( B )}{cos ^{2} ( B )}

\frac{1}{cos ( B )} \cdot \frac{sin ( B )}{cos ( B )}

Multiply fractions:

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

= \frac{1 \cdot sin ( B )}{cos ( B ) cos ( B )}

Multiply:

1 \cdot sin (B) = sin (B)

= \frac{sin ( B )}{cos ( B ) cos ( B )}

cos (B) cos (B) = cos^{2} (B)

cos (B) cos (B)

Apply exponent rule:

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

cos (B) cos (B) = cos^{1 + 1} (B)

= cos^{1 + 1} (B)

Add the numbers:

1 + 1 = 2

= cos^{2} (B)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Factor

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

1 - sin^{2} (B)

Apply Difference of Two Squares Formula:

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

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

= (1 + sin (B)) (1 - sin (B))

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

Cancel the common factor:

1 + sin (B)

= \frac{1}{1 - sin ( B )}

= \frac{1}{1 - sin ( B )}

We showed that the two sides could take the same form

\Rightarrow True

tan (6 0^{\circ}) = \frac{2 tan ( 3 0 ^{\circ} )}{1 - tan ^{2} ( 3 0 ^{\circ} )}

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

cos (x) + sin (x) = 1

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

sin^{2} (u) (cot^{2} (u) + 1) = 1

Is 1/(1-sin(B))=sec^2(B)+sec(B)tan(B) ?
The answer to whether 1/(1-sin(B))=sec^2(B)+sec(B)tan(B) is True