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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Least Common Multiplier of

1 + sin (x), cos (x) : cos (x) (sin (x) + 1)

1 + sin (x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (x)

cos (x)

= cos (x) (sin (x) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (x) (sin (x) + 1)

For

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

multiply the denominator and numerator by

cos (x)

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

For

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

multiply the denominator and numerator by

sin (x) + 1

\frac{1}{cos ( x )} = \frac{1 \cdot ( sin ( x ) + 1 )}{cos ( x ) ( sin ( x ) + 1 )} = \frac{sin ( x ) + 1}{cos ( x ) ( sin ( x ) + 1 )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

sin^{2} (x) + sin (x) : sin (x) (sin (x) + 1)

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

Apply exponent rule:

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

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

= sin (x) sin (x) + sin (x)

Factor out common term

sin (x)

= sin (x) (sin (x) + 1)

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

Cancel the common factor:

sin (x) + 1

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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