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

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

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prove ((1+cot^2(x)))/(sec^2(x))=cot^2(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{1 + cot ^{2} ( x )}{sec ^{2} ( x )}

Express with sin, cos

\frac{1 + cot ^{2} ( x )}{sec ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

cot (x) cot (x)

Simplify

cot (x) cot (x) : cot^{2} (x)

cot (x) cot (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cot (x) cot (x) = cot^{1 + 1} (x)

= cot^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cot^{2} (x)

cot^{2} (x)

cot^{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+cot^2(x)))/(sec^2(x))=cot^2(x) ?
The answer to whether ((1+cot^2(x)))/(sec^2(x))=cot^2(x) is True