Is (1+sin(2x))/(sin(2x))=1+1/2 sec(x)csc(x) ?

The answer to whether (1+sin(2x))/(sin(2x))=1+1/2 sec(x)csc(x) is True

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prove (1+sin(2x))/(sin(2x))=1+1/2 sec(x)csc(x)

prove

\frac{1 + sin ( 2 x )}{sin ( 2 x )} = 1 + \frac{1}{2} sec (x) csc (x)

True

Solution steps

\frac{1 + sin ( 2 x )}{sin ( 2 x )} = 1 + \frac{1}{2} sec (x) csc (x)

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Manipulating right side

1 + \frac{1}{2} sec (x) csc (x)

Express with sin, cos

1 + csc (x) \frac{1}{2} sec (x)

Use the basic trigonometric identity:

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

= 1 + \frac{1}{sin ( x )} \cdot \frac{1}{2} sec (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

1 + \frac{1}{sin ( x )} \cdot \frac{1}{2} \cdot \frac{1}{cos ( x )} : \frac{2 sin ( x ) cos ( x ) + 1}{2 sin ( x ) cos ( x )}

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 1 \cdot 1 = 1

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{1}{csc ( x ) + 1} + \frac{1}{sin ( x ) + 1} = 1

p ro v e cos (2 u) = cos^{2} (u) - sin^{2} (u)

p ro v e (csc (θ) + 1) (csc (θ) - 1) = cot^{2} (θ)

p ro v e cos (θ) (cot (θ) + tan (θ)) = csc (θ)

p ro v e sin (- θ) = - sin (θ)

Is (1+sin(2x))/(sin(2x))=1+1/2 sec(x)csc(x) ?
The answer to whether (1+sin(2x))/(sin(2x))=1+1/2 sec(x)csc(x) is True