Is cot(2θ)=(csc(θ)-2sin(θ))/(2cos(θ)) ?

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

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prove cot(2θ)=(csc(θ)-2sin(θ))/(2cos(θ))

prove

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

True

Solution steps

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

Manipulating right side

\frac{csc ( θ ) - 2 sin ( θ )}{2 cos ( θ )}

Express with sin, cos

\frac{csc ( θ ) - 2 sin ( θ )}{2 cos ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( θ )} - 2 sin ( θ )}{2 cos ( θ )}

Simplify

\frac{\frac{1}{s i n ( θ )} - 2 sin ( θ )}{2 cos ( θ )} : \frac{1 - 2 sin ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

\frac{\frac{1}{s i n ( θ )} - 2 sin ( θ )}{2 cos ( θ )}

Join

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

\frac{1}{sin ( θ )} - 2 sin (θ)

Convert element to fraction:

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

= \frac{1}{sin ( θ )} - \frac{2 sin ( θ ) sin ( θ )}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 - 2 sin ( θ ) sin ( θ )}{sin ( θ )}

1 - 2 sin (θ) sin (θ) = 1 - 2 sin^{2} (θ)

1 - 2 sin (θ) sin (θ)

2 sin (θ) sin (θ) = 2 sin^{2} (θ)

2 sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 sin^{2} (θ)

= 1 - 2 sin^{2} (θ)

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

= \frac{cos ( 2 θ )}{sin ( 2 θ )}

Use the basic trigonometric identity:

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

= cot (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin^{2} (θ) sec^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

p ro v e \frac{csc ^{2} ( x )}{cot ( x )} = sec (x) csc (x)

p ro v e \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )} = 2 sec (A)

p ro v e cot (\frac{x}{2}) = \frac{sin ( x )}{1 - cos ( x )}

p ro v e - tan (y) + sec (y) = \frac{cos ( y )}{1 + sin ( y )}

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