Is sec^2(θ)-1=(tan(θ))/(cot(θ)) ?

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

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Popular Trigonometry >

prove sec^2(θ)-1=(tan(θ))/(cot(θ))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

sec^{2} (θ) - 1

Express with sin, cos

- 1 + sec^{2} (θ)

Use the basic trigonometric identity:

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

= - 1 + (\frac{1}{cos ( θ )})^{2}

Simplify

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

- 1 + (\frac{1}{cos ( θ )})^{2}

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= - 1 + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

\frac{tan ( θ )}{cot ( θ )}

Express with sin, cos

\frac{tan ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( θ )}{c o s ( θ )}}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

Simplify

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

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

Divide fractions:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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