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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

cot (x) cos (2 x) = - sin (2 x) + cot (x)

cot (u) tan (u) - cos^{2} (u) = sin^{2} (u)

\frac{csc ^{4} ( x ) - 1}{cot ^{2} ( x )} = csc^{2} (x) + 1

sin (2 x) tan (x) = 1 - cos (2 x)

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