Is sec(2x)=(csc(x))/(csc(x)-2sin(x)) ?

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

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Popular >

prove sec(2x)=(csc(x))/(csc(x)-2sin(x))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (2 x)

sec (2 x)

sec (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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