Is tan(2a)-2tan(2a)sin^2(a)=sin(2a) ?

The answer to whether tan(2a)-2tan(2a)sin^2(a)=sin(2a) is True

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prove tan(2a)-2tan(2a)sin^2(a)=sin(2a)

prove

tan (2 a) - 2 tan (2 a) sin^{2} (a) = sin (2 a)

True

Solution steps

tan (2 a) - 2 tan (2 a) sin^{2} (a) = sin (2 a)

Manipulating left side

tan (2 a) - 2 tan (2 a) sin^{2} (a)

Express with sin, cos

tan (2 a) - 2 sin^{2} (a) tan (2 a)

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{sin ( 2 a )}{cos ( 2 a )} - 2 sin^{2} (a) \frac{sin ( 2 a )}{cos ( 2 a )}

Multiply

2 sin^{2} (a) \frac{sin ( 2 a )}{cos ( 2 a )} : \frac{2 sin ^{2} ( a ) sin ( 2 a )}{cos ( 2 a )}

2 sin^{2} (a) \frac{sin ( 2 a )}{cos ( 2 a )}

Multiply fractions:

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

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{sin ( 2 a ) - 2 sin ( 2 a ) sin ^{2} ( a )}{cos ( 2 a )}

Use the Double Angle identity:

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

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

= \frac{sin ( 2 a ) + ( cos ( 2 a ) - 1 ) sin ( 2 a )}{cos ( 2 a )}

Simplify

\frac{sin ( 2 a ) + ( cos ( 2 a ) - 1 ) sin ( 2 a )}{cos ( 2 a )} : sin (2 a)

\frac{sin ( 2 a ) + ( cos ( 2 a ) - 1 ) sin ( 2 a )}{cos ( 2 a )}

Factor

sin (2 a) + (cos (2 a) - 1) sin (2 a) : sin (2 a) cos (2 a)

sin (2 a) + (cos (2 a) - 1) sin (2 a)

Factor out common term

sin (2 a)

= sin (2 a) (1 - 1 + cos (2 a))

Refine

= sin (2 a) cos (2 a)

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

Cancel the common factor:

cos (2 a)

= sin (2 a)

= sin (2 a)

= sin (2 a)

We showed that the two sides could take the same form

\Rightarrow True

tan (t) sin (t) = \frac{sec ^{2} ( t ) - 1}{sec ( t )}

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

80 = arcsin (45.6443 \cdot \frac{sin ( c )}{35})

sin (0) csc (0) = 1

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

Is tan(2a)-2tan(2a)sin^2(a)=sin(2a) ?
The answer to whether tan(2a)-2tan(2a)sin^2(a)=sin(2a) is True