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

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

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Popular Trigonometry >

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

Solution

prove

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{2 ( \frac{1}{c o s ( x )} ) ^{2} - 2 tan ^{2} ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

= \frac{2 ( \frac{1}{c o s ( x )} ) ^{2} - 2 ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{csc ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{2 ( \frac{1}{c o s ( x )} ) ^{2} - 2 ( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{\frac{1}{s i n ( x )}}

Apply the fraction rule:

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

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

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

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

Combine the fractions

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

Apply rule

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

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

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

Remove parentheses:

(a) = a

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

Multiply

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

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

Multiply fractions:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Factor

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

2 - 2 sin^{2} (x)

Rewrite as

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

Factor out common term

2

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

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

Cancel the common factor:

1 - sin^{2} (x)

= 2 sin (x)

= 2 sin (x)

= 2 sin (x)

Manipulating right side

sin (2 x) sec (x)

Express with sin, cos

sec (x) sin (2 x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (2 x) = sin (2 x)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Cancel the common factor:

cos (x)

= 2 sin (x)

= 2 sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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