Is cot(θ)(tan(θ)+cot(θ))=csc^2(θ) ?

The answer to whether cot(θ)(tan(θ)+cot(θ))=csc^2(θ) is True

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prove cot(θ)(tan(θ)+cot(θ))=csc^2(θ)

Solution

prove

cot (θ) (tan (θ) + cot (θ)) = csc^{2} (θ)

Solution

True

Solution steps

cot (θ) (tan (θ) + cot (θ)) = csc^{2} (θ)

Manipulating left side

cot (θ) (tan (θ) + cot (θ))

Express with sin, cos

(cot (θ) + tan (θ)) cot (θ)

Use the basic trigonometric identity:

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

= (\frac{cos ( θ )}{sin ( θ )} + tan (θ)) \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{cos ( θ )}{sin ( θ )} + \frac{sin ( θ )}{cos ( θ )}) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

= \frac{cos ( θ ) ( \frac{c o s ( θ )}{s i n ( θ )} + \frac{s i n ( θ )}{c o s ( θ )} )}{sin ( θ )}

Join

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

\frac{cos ( θ )}{sin ( θ )} + \frac{sin ( θ )}{cos ( θ )}

Least Common Multiplier of

sin (θ), cos (θ) : sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (θ) cos (θ)

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{s i n ( θ ) c o s ( θ )} cos ( θ )}{sin ( θ )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

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

= \frac{\frac{c o s ^{2} ( θ ) + 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{cos ^{2} ( θ ) + 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{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( θ )} ) ^{2}}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= \frac{1}{\frac{1}{c s c ^{2} ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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