Is (cos(2x))/(sin(x))=cos(x)cot(x)-sin(x) ?

The answer to whether (cos(2x))/(sin(x))=cos(x)cot(x)-sin(x) is True

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prove (cos(2x))/(sin(x))=cos(x)cot(x)-sin(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Manipulating right side

cos (x) cot (x) - sin (x)

Express with sin, cos

- sin (x) + cos (x) cot (x)

Use the basic trigonometric identity:

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

= - sin (x) + cos (x) \frac{cos ( x )}{sin ( x )}

Simplify

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

- sin (x) + cos (x) \frac{cos ( x )}{sin ( x )}

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

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

Multiply fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (2 x) + cos^{2} (2 x) + sin^{2} (2 x) = csc^{2} (2 x)

sin^{2} (x) - 4 cos^{2} (x) = 1 - 5 cos^{2} (x)

25 sec^{2} (5 x) = 25 + (5 tan (5 x))^{2}

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

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

Is (cos(2x))/(sin(x))=cos(x)cot(x)-sin(x) ?
The answer to whether (cos(2x))/(sin(x))=cos(x)cot(x)-sin(x) is True