Is (2cot(u))/(csc^2(u)-2)=tan(2u) ?

The answer to whether (2cot(u))/(csc^2(u)-2)=tan(2u) is True

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prove (2cot(u))/(csc^2(u)-2)=tan(2u)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{2 cot ( u )}{csc ^{2} ( u ) - 2}

Express with sin, cos

\frac{2 cot ( u )}{- 2 + csc ^{2} ( u )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( u )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply

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

2 \cdot \frac{cos ( u )}{sin ( u )}

Multiply fractions:

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

= \frac{cos ( u ) \cdot 2}{sin ( u )}

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

2 = \frac{2 sin ^{2} ( u )}{sin ^{2} ( u )}

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{2 cos ( u )}{\frac{- 2 s i n ^{2} ( u ) + 1}{s i n ^{2} ( u )} sin ( u )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (u)

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

= \frac{2 cos ( u )}{\frac{- 2 s i n ^{2} ( u ) + 1}{s i n ( u )}}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

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

Use the basic trigonometric identity:

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

= tan (2 u)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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