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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{tan ( θ )}{1 + sec ( θ )}

Express with sin, cos

\frac{tan ( θ )}{1 + sec ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= cos (θ) + 1

= \frac{sin ( θ )}{cos ( θ ) + 1}

= \frac{sin ( θ )}{cos ( θ ) + 1}

= \frac{sin ( θ )}{1 + cos ( θ )}

Manipulating right side

- cot (θ) + csc (θ)

Express with sin, cos

- cot (θ) + csc (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

= \frac{- cos ( θ ) + 1}{sin ( θ )}

= \frac{- cos ( θ ) + 1}{sin ( θ )}

= \frac{1 - cos ( θ )}{sin ( θ )}

Rewrite using trig identities

Multiply by

\frac{1 + cos ( θ )}{1 + cos ( θ )}

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

Expand

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

(1 + cos (θ)) (1 - cos (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = cos (θ)

= 1^{2} - cos^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (θ)

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

Use the Pythagorean identity:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - cos^{2} (θ) = sin^{2} (θ)

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

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{1 + cos ( θ )}

= \frac{sin ( θ )}{1 + cos ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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