Is (sec(θ)+tan(θ))(1-sin(θ))=cos(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

(sec (θ) + tan (θ)) (1 - sin (θ))

Express with sin, cos

(1 - sin (θ)) (sec (θ) + tan (θ))

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Simplify

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

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

Apply rule

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

= \frac{1 + sin ( θ )}{cos ( θ )}

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sin (arcsin (x) + arccos (x)) = 1

\frac{1 - 2 sec ( x ) - 3 sec ^{2} ( x )}{- tan ^{2} ( x )} = \frac{1 - 3 sec ( x )}{1 - sec ( x )}

\frac{1}{2 csc ( 2 x )} = cos^{2} (x) tan (x)

(sec (5 x) - 1) (sec (5 x) - 1) = tan^{2} (5 x)

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