Is (2cos(x))/(csc(x)-2sin(x))=tan(2x) ?

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

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prove (2cos(x))/(csc(x)-2sin(x))=tan(2x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - 2 sin (x) sin (x)

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

2 sin (x) sin (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

= 2 sin^{2} (x)

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

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

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

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

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

Use the basic trigonometric identity:

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

= tan (2 x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + csc ( a )}{cot ( a ) + cos ( a )} = sec (a)

sec (θ) - sec (θ) sin^{2} (θ) = cos (θ)

- 2 = - 2 cos (0) + 15 sin (3) (0)

tan (x) - sin (x) = tan (x) sin (x)

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

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