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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Divide fractions:

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

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

Refine

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

Cancel the common factor:

sin^{2} (x)

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

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

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

Manipulating right side

1 + tan^{2} (x)

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{1}{cos ^{2} ( 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))/(cot^2(x))=1+tan^2(x) ?
The answer to whether (csc^2(x))/(cot^2(x))=1+tan^2(x) is True