Is csc^2(x)tan(x)-cot(x)=tan(x) ?

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

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Popular Trigonometry >

prove csc^2(x)tan(x)-cot(x)=tan(x)

Solution

prove

csc^{2} (x) tan (x) - cot (x) = tan (x)

Solution

True

Solution steps

csc^{2} (x) tan (x) - cot (x) = tan (x)

Manipulating left side

csc^{2} (x) tan (x) - cot (x)

Express with sin, cos

- cot (x) + csc^{2} (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

Cancel the common factor:

sin (x)

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

= \frac{\frac{1}{s i n ( x )}}{cos ( x )}

Apply the fraction rule:

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

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

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

Least Common Multiplier of

sin (x), sin (x) cos (x) : sin (x) cos (x)

sin (x), sin (x) cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (x)

sin (x) cos (x)

= sin (x) cos (x)

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 (x) cos (x)

For

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

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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