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

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

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

prove

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

True

Solution steps

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

Manipulating left side

\frac{csc ^{2} ( x )}{2 cot ( x )}

Express with sin, cos

\frac{csc ^{2} ( x )}{2 cot ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin (x)

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

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

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

Manipulating right side

csc (2 x)

Express with sin, cos

csc (2 x)

Use the basic trigonometric identity:

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cot (x) (sec^{2} (x) - 1) = tan (x)

p ro v e (1 - cos (x)) (1 + cos (x)) = sin^{2} (x)

p ro v e cot (2 θ) = \frac{cot ^{2} ( θ ) - 1}{2 cot ( θ )}

p ro v e (sec^{2} (x) - 1) cos (x) = sin^{2} (x) sec (x)

p ro v e \frac{2}{cot ( x ) + tan ( x )} = sin (2 x)

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