Is (tan(β)+sec(β))(1-sin(β))=cos(β) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

(tan (β) + sec (β)) (1 - sin (β))

Express with sin, cos

(1 - sin (β)) (sec (β) + tan (β))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Apply rule

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

= \frac{1 + sin ( β )}{cos ( β )}

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (β)) (1 - sin (β))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (β)

= 1^{2} - sin^{2} (β)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (β)

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

Use the Pythagorean identity:

1 = cos^{2} (β) + sin^{2} (β)

1 - sin^{2} (β) = cos^{2} (β)

= \frac{cos ^{2} ( β )}{cos ( β )}

Cancel the common factor:

cos (β)

= cos (β)

= cos (β)

We showed that the two sides could take the same form

\Rightarrow True

sin (α + β) = sin (α) cos (β) + sin (β) cos (α)

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

2 - sec^{2} (z) = 1 - tan^{2} (z)

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

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

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