Is 2csc(2x)=csc^2(x)tan(x) ?

The answer to whether 2csc(2x)=csc^2(x)tan(x) is True

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prove 2csc(2x)=csc^2(x)tan(x)

Solution

prove

2 csc (2 x) = csc^{2} (x) tan (x)

Solution

True

Solution steps

2 csc (2 x) = csc^{2} (x) tan (x)

Manipulating left side

2 csc (2 x)

Express with sin, cos

2 csc (2 x)

Use the basic trigonometric identity:

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

= 2 \cdot \frac{1}{sin ( 2 x )}

Simplify

2 \cdot \frac{1}{sin ( 2 x )} : \frac{2}{sin ( 2 x )}

2 \cdot \frac{1}{sin ( 2 x )}

Multiply fractions:

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

= \frac{1 \cdot 2}{sin ( 2 x )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sin ( 2 x )}

= \frac{2}{sin ( 2 x )}

= \frac{2}{sin ( 2 x )}

Rewrite using trig identities

\frac{2}{sin ( 2 x )}

Use the Double Angle identity:

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

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

Divide the numbers:

\frac{2}{2} = 1

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

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

Manipulating right side

csc^{2} (x) tan (x)

Express with sin, cos

csc^{2} (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

= \frac{sin ( x ) ( \frac{1}{s i n ( x )} ) ^{2}}{cos ( x )}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{\frac{1}{s i n ^{2} ( x )} sin ( x )}{cos ( x )}

Multiply

sin (x) \frac{1}{sin ^{2} ( x )} : \frac{1}{sin ( x )}

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

Cancel the common factor:

sin (x)

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

= \frac{\frac{1}{s i n ( x )}}{cos ( x )}

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 2csc(2x)=csc^2(x)tan(x) ?
The answer to whether 2csc(2x)=csc^2(x)tan(x) is True