Is tan(2θ)=(2cot(θ))/(cot^2(θ)-1) ?

The answer to whether tan(2θ)=(2cot(θ))/(cot^2(θ)-1) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan(2θ)=(2cot(θ))/(cot^2(θ)-1)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

tan (2 θ)

Express with sin, cos

tan (2 θ)

Use the basic trigonometric identity:

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

= \frac{sin ( 2 θ )}{cos ( 2 θ )}

= \frac{sin ( 2 θ )}{cos ( 2 θ )}

Rewrite using trig identities

\frac{sin ( 2 θ )}{cos ( 2 θ )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{2 sin ( θ ) cos ( θ )}{cos ( 2 θ )}

Use the Double Angle identity:

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

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

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

Manipulating right side

\frac{2 cot ( θ )}{cot ^{2} ( θ ) - 1}

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{2 \cdot \frac{c o s ( θ )}{s i n ( θ )}}{- 1 + ( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}

Simplify

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

\frac{2 \cdot \frac{c o s ( θ )}{s i n ( θ )}}{- 1 + ( \frac{c o s ( θ )}{s i n ( θ )} ) ^{2}}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 \cdot \frac{c o s ( θ )}{s i n ( θ )}}{- 1 + \frac{c o s ^{2} ( θ )}{s i n ^{2} ( θ )}}

Multiply

2 \cdot \frac{cos ( θ )}{sin ( θ )} : \frac{2 cos ( θ )}{sin ( θ )}

2 \cdot \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

= \frac{cos ( θ ) \cdot 2}{sin ( θ )}

= \frac{\frac{2 c o s ( θ )}{s i n ( θ )}}{- 1 + \frac{c o s ^{2} ( θ )}{s i n ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{cos ( θ ) \cdot 2}{sin ( θ ) ( - 1 + \frac{c o s ^{2} ( θ )}{s i n ^{2} ( θ )} )}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

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

= \frac{2 cos ( θ )}{\frac{- s i n ^{2} ( θ ) + c o s ^{2} ( θ )}{s i n ( θ )}}

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove csc(x)-tan(x/2)=cot(x)prove

csc (x) - tan (\frac{x}{2}) = cot (x)

prove cos(3x)-cos(x)=4cos^3(x)-4cos(x)prove

cos (3 x) - cos (x) = 4 cos^{3} (x) - 4 cos (x)

prove cot^2(θ)-cos^2(θ)=cos^2(θ)cot^2(θ)prove

cot^{2} (θ) - cos^{2} (θ) = cos^{2} (θ) cot^{2} (θ)

prove (tan(A))/(sec(A))=sin(A)prove

\frac{tan ( A )}{sec ( A )} = sin (A)

prove tan(8x)-tan(8x)tan^2(4x)=2tan(4x)prove

tan (8 x) - tan (8 x) tan^{2} (4 x) = 2 tan (4 x)

Frequently Asked Questions (FAQ)

Is tan(2θ)=(2cot(θ))/(cot^2(θ)-1) ?
The answer to whether tan(2θ)=(2cot(θ))/(cot^2(θ)-1) is True