Is (tan(θ)+cot(θ))/(tan(θ))=csc^2(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (tan(θ)+cot(θ))/(tan(θ))=csc^2(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{tan ( θ ) + cot ( θ )}{tan ( θ )}

Express with sin, cos

\frac{cot ( θ ) + tan ( θ )}{tan ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Popular Examples

prove sin(60)=2sin(30)cos(30)prove

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

prove (cos(θ))/(1+sin(θ))+tan(θ)=sec(θ)prove

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ) = sec (θ)

prove cos(x)*tan(x)=sin(x)prove

cos (x) \cdot tan (x) = sin (x)

prove (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ)prove

\frac{sin ( 2 θ )}{sin ( θ )} - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)

prove (sec(x))/(csc(x))+(sin(x))/(cos(x))=2tan(x)prove

\frac{sec ( x )}{csc ( x )} + \frac{sin ( x )}{cos ( x )} = 2 tan (x)

Frequently Asked Questions (FAQ)

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