Is (tan(θ))/(csc(θ))=sec(θ)-cos(θ) ?

The answer to whether (tan(θ))/(csc(θ))=sec(θ)-cos(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (tan(θ))/(csc(θ))=sec(θ)-cos(θ)

prove

\frac{tan ( θ )}{csc ( θ )} = sec (θ) - cos (θ)

True

Solution steps

\frac{tan ( θ )}{csc ( θ )} = sec (θ) - cos (θ)

Manipulating left side

\frac{tan ( θ )}{csc ( θ )}

Express with sin, cos

\frac{tan ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{1}{s i n ( θ )}} : \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Divide fractions:

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

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

Refine

= \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 )}

Manipulating right side

sec (θ) - cos (θ)

Express with sin, cos

- cos (θ) + sec (θ)

Use the basic trigonometric identity:

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

= - cos (θ) + \frac{1}{cos ( θ )}

Simplify

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

We showed that the two sides could take the same form

\Rightarrow True

cot (2 β) = \frac{cot ^{2} ( β ) - 1}{2 cot ( β )}

(2 sin (x) - 2 cos (x))^{2} = 4 - 4 sin (2 x)

csc^{2} (y) tan^{2} (y) - 1 = tan^{2} (y)

cot (θ) tan (θ) = 1

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

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