Is 1/(1-sin(θ))=sec^2(θ)+sec(θ)tan(θ) ?

The answer to whether 1/(1-sin(θ))=sec^2(θ)+sec(θ)tan(θ) is True

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prove 1/(1-sin(θ))=sec^2(θ)+sec(θ)tan(θ)

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (θ) + sec (θ) tan (θ)

Express with sin, cos

sec^{2} (θ) + sec (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

(\frac{1}{cos ( θ )})^{2} + \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{1 + sin ( θ )}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2} + \frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ( θ ) cos ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 + sin ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

cos^{2} (θ) = 1 - sin^{2} (θ)

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

Factor

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

1 - sin^{2} (θ)

Apply Difference of Two Squares Formula:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

= (1 + sin (θ)) (1 - sin (θ))

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

Cancel the common factor:

1 + sin (θ)

= \frac{1}{1 - sin ( θ )}

= \frac{1}{1 - sin ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

Is 1/(1-sin(θ))=sec^2(θ)+sec(θ)tan(θ) ?
The answer to whether 1/(1-sin(θ))=sec^2(θ)+sec(θ)tan(θ) is True