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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

sec^{2} (x)

Express with sin, cos

sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

(\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 )}

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

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

Express with sin, cos

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

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

1 - sin (x) sin (x)

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= 1 - sin^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

= 1 - sin^{2} (x)

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

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

= \frac{1}{1 - sin ^{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^2(x)=(csc(x))/(csc(x)-sin(x)) ?
The answer to whether sec^2(x)=(csc(x))/(csc(x)-sin(x)) is True