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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (b) + tan (b)

Express with sin, cos

sec (b) + tan (b)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( b )} + tan (b)

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1}{cos ( b )} + \frac{sin ( b )}{cos ( b )}

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

\frac{1 - sin ( b )}{1 - sin ( b )}

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

Expand

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

(1 + sin (b)) (1 - sin (b))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (b)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (b)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (b)

= \frac{cos ( b )}{1 - sin ( b )}

= \frac{cos ( b )}{1 - sin ( b )}

We showed that the two sides could take the same form

\Rightarrow True

cot^{4} (x) + cot^{2} (x) = csc^{4} (x) - csc^{2} (x)

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

cos (θ) cot (θ) = csc (θ) - sin (θ)

1 - tanh^{2} (x) = sech^{2} (x)

tan (\frac{x}{2}) - cot (\frac{x}{2}) = - 2 cot (x)

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