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

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

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prove 1+cot^2(θ)=(sec^2(θ))/(sec^2(θ)-1)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

1 + cot^{2} (θ)

Express with sin, cos

1 + cot^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

1 + (\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

= \frac{sin ^{2} ( θ ) + cos ^{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

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

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

cos^{2} (θ) + 1 - cos^{2} (θ) = 1

cos^{2} (θ) + 1 - cos^{2} (θ)

Group like terms

= cos^{2} (θ) - cos^{2} (θ) + 1

Add similar elements:

cos^{2} (θ) - cos^{2} (θ) = 0

= 1

= \frac{1}{1 - cos ^{2} ( θ )}

= \frac{1}{1 - cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{1 - ( \frac{1}{s e c ( θ )} ) ^{2}}

Simplify

\frac{1}{1 - ( \frac{1}{s e c ( θ )} ) ^{2}}

(\frac{1}{sec ( θ )})^{2} = \frac{1}{sec ^{2} ( θ )}

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

= \frac{1}{1 - \frac{1}{s e c ^{2} ( θ )}}

Join

1 - \frac{1}{sec ^{2} ( θ )} : \frac{sec ^{2} ( θ ) - 1}{sec ^{2} ( θ )}

1 - \frac{1}{sec ^{2} ( θ )}

Convert element to fraction:

1 = \frac{1 sec ^{2} ( θ )}{sec ^{2} ( θ )}

= \frac{1 \cdot sec ^{2} ( θ )}{sec ^{2} ( θ )} - \frac{1}{sec ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sec ^{2} ( θ ) - 1}{sec ^{2} ( θ )}

Multiply:

1 \cdot sec^{2} (θ) = sec^{2} (θ)

= \frac{sec ^{2} ( θ ) - 1}{sec ^{2} ( θ )}

= \frac{1}{\frac{s e c ^{2} ( θ ) - 1}{s e c ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{sec ^{2} ( θ )}{sec ^{2} ( θ ) - 1}

\frac{sec ^{2} ( θ )}{sec ^{2} ( θ ) - 1}

\frac{sec ^{2} ( θ )}{sec ^{2} ( θ ) - 1}

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