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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

csc (x) - sin (x)

Express with sin, cos

csc (x) - sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

\frac{cot ^{2} ( x )}{csc ( x )}

Express with sin, cos

\frac{cot ^{2} ( x )}{csc ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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