Is (tan(x))/(1+sec(x))=-cot(x)+csc(x) ?

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

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prove (tan(x))/(1+sec(x))=-cot(x)+csc(x)

Solution

prove

\frac{tan ( x )}{1 + sec ( x )} = - cot (x) + csc (x)

Solution

True

Solution steps

\frac{tan ( x )}{1 + sec ( x )} = - cot (x) + csc (x)

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{1 + sec ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( x )}{c o s ( x )}}{1 + \frac{1}{c o s ( x )}}

Simplify

\frac{\frac{s i n ( x )}{c o s ( x )}}{1 + \frac{1}{c o s ( x )}} : \frac{sin ( x )}{cos ( x ) + 1}

\frac{\frac{s i n ( x )}{c o s ( x )}}{1 + \frac{1}{c o s ( x )}}

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= cos (x) + 1

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

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

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

Manipulating right side

- cot (x) + csc (x)

Express with sin, cos

- cot (x) + csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

Multiply by

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

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

Expand

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

(1 + cos (x)) (1 - cos (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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