Is csc(x)sec(x)=2csc(2x) ?

The answer to whether csc(x)sec(x)=2csc(2x) is True

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prove csc(x)sec(x)=2csc(2x)

prove

csc (x) sec (x) = 2 csc (2 x)

True

Solution steps

csc (x) sec (x) = 2 csc (2 x)

Manipulating left side

csc (x) sec (x)

Express with sin, cos

csc (x) sec (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{2}{\frac{1}{c s c ( 2 x )}}

Simplify

\frac{2}{\frac{1}{c s c ( 2 x )}}

Apply the fraction rule:

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

= \frac{2 csc ( 2 x )}{1}

Apply rule

\frac{a}{1} = a

= 2 csc (2 x)

2 csc (2 x)

2 csc (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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

2 sec^{2} (x) - 2 cot^{2} (\frac{π}{2} - x) = 2

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

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

sin^{2} (θ) - 2 = - 1 - cos^{2} (θ)