Is sec(θ)-1/(sec(θ))=sin(θ)tan(θ) ?

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

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prove sec(θ)-1/(sec(θ))=sin(θ)tan(θ)

prove

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

True

Solution steps

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

Manipulating left side

sec (θ) - \frac{1}{sec ( θ )}

Express with sin, cos

- \frac{1}{sec ( θ )} + sec (θ)

Use the basic trigonometric identity:

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

= - \frac{1}{\frac{1}{c o s ( θ )}} + \frac{1}{cos ( θ )}

Simplify

- \frac{1}{\frac{1}{c o s ( θ )}} + \frac{1}{cos ( θ )} : \frac{- cos ^{2} ( θ ) + 1}{cos ( θ )}

- \frac{1}{\frac{1}{c o s ( θ )}} + \frac{1}{cos ( θ )}

\frac{1}{\frac{1}{c o s ( θ )}} = cos (θ)

\frac{1}{\frac{1}{c o s ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{cos ( θ )}{1}

Apply rule

\frac{a}{1} = a

= cos (θ)

= - cos (θ) + \frac{1}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) cos (θ) + 1

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)

= - cos^{2} (θ) + 1

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

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

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

Manipulating right side

sin (θ) tan (θ)

Express with sin, cos

sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (θ) + sec^{2} (θ) = csc^{2} (θ) sec^{2} (θ)

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

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

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

\frac{sin ( 2 x )}{2} = \frac{1}{2} sin (2 x)

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