Is (1+tan^2(x))cot^2(x)=csc^2(x) ?

The answer to whether (1+tan^2(x))cot^2(x)=csc^2(x) is True

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Popular >

prove (1+tan^2(x))cot^2(x)=csc^2(x)

Solution

prove

(1 + tan^{2} (x)) cot^{2} (x) = csc^{2} (x)

Solution

True

Solution steps

(1 + tan^{2} (x)) cot^{2} (x) = csc^{2} (x)

Manipulating left side

(1 + tan^{2} (x)) cot^{2} (x)

Express with sin, cos

(1 + tan^{2} (x)) cot^{2} (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc^{2} (x)

csc^{2} (x)

csc^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (1+tan^2(x))cot^2(x)=csc^2(x) ?
The answer to whether (1+tan^2(x))cot^2(x)=csc^2(x) is True