Is csc(θ)(csc(θ)-sin(θ))=cot^2(θ) ?

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

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prove csc(θ)(csc(θ)-sin(θ))=cot^2(θ)

Solution

prove

csc (θ) (csc (θ) - sin (θ)) = cot^{2} (θ)

Solution

True

Solution steps

csc (θ) (csc (θ) - sin (θ)) = cot^{2} (θ)

Manipulating left side

csc (θ) (csc (θ) - sin (θ))

Express with sin, cos

(csc (θ) - sin (θ)) csc (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

= (\frac{1}{sin ( θ )} - sin (θ))

Remove parentheses:

(a) = a

= \frac{1}{sin ( θ )} - sin (θ)

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

Join

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= 1 - sin^{2} (θ)

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

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

Apply the fraction rule:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

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

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

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

Manipulating right side

cot^{2} (θ)

Express with sin, cos

cot^{2} (θ)

Use the basic trigonometric identity:

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

= (\frac{cos ( θ )}{sin ( θ )})^{2}

Simplify

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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