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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

csc (x) (csc (x) + sin (- x))

Use the negative angle identity:

sin (- x) = - sin (x)

= csc (x) (csc (x) - sin (x))

Express with sin, cos

(csc (x) - sin (x)) csc (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

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

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

Apply the fraction rule:

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

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

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

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

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

Manipulating right side

cot^{2} (x)

Express with sin, cos

cot^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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