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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

\frac{( 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 )}{cos ( b )}

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (b)

= cos (b)

= cos (b)

We showed that the two sides could take the same form

\Rightarrow True

cot (2 x) = \frac{cos ( 2 x )}{sin ( 2 x )}

cos^{2} (a) + cos^{2} (a) cot^{2} (a) = cot^{2} (a)

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

\frac{csc ( a )}{cot ( a )} = sec (a)

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

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