Is sec(x)-1/(sec(x))=sin(x)tan(x) ?

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

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prove sec(x)-1/(sec(x))=sin(x)tan(x)

prove

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

True

Solution steps

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

Manipulating left side

sec (x) - \frac{1}{sec ( x )}

Express with sin, cos

- \frac{1}{sec ( x )} + sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{\frac{1}{c o s ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= cos (x)

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (x) cos (x) + 1

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

Manipulating right side

sin (x) tan (x)

Express with sin, cos

sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= \frac{sin ^{2} ( x )}{cos ( x )}

= \frac{sin ^{2} ( x )}{cos ( x )}

= \frac{sin ^{2} ( x )}{cos ( x )}

Rewrite using trig identities

\frac{sin ^{2} ( x )}{cos ( x )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )} = sec (b)

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

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

tan (θ) = sin (θ) sec (θ)

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

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