Is (1+cot^2(x))/(2cot(x))=csc(2x) ?

The answer to whether (1+cot^2(x))/(2cot(x))=csc(2x) is True

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prove (1+cot^2(x))/(2cot(x))=csc(2x)

Solution

prove

\frac{1 + cot ^{2} ( x )}{2 cot ( x )} = csc (2 x)

Solution

True

Solution steps

\frac{1 + cot ^{2} ( x )}{2 cot ( x )} = csc (2 x)

Manipulating left side

\frac{1 + cot ^{2} ( x )}{2 cot ( x )}

Express with sin, cos

\frac{1 + cot ^{2} ( x )}{2 cot ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( x ) + cos ^{2} ( x )}{sin ^{2} ( x )}

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( 2 x )}}

Simplify

\frac{1}{\frac{1}{c s c ( 2 x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( 2 x )}{1}

Apply rule

\frac{a}{1} = a

= csc (2 x)

csc (2 x)

csc (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (1+cot^2(x))/(2cot(x))=csc(2x) ?
The answer to whether (1+cot^2(x))/(2cot(x))=csc(2x) is True