Is (1+cot^2(θ))tan^2(θ)=sec^2(θ) ?

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

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prove (1+cot^2(θ))tan^2(θ)=sec^2(θ)

Solution

prove

(1 + cot^{2} (θ)) tan^{2} (θ) = sec^{2} (θ)

Solution

True

Solution steps

(1 + cot^{2} (θ)) tan^{2} (θ) = sec^{2} (θ)

Manipulating left side

(1 + cot^{2} (θ)) tan^{2} (θ)

Express with sin, cos

(1 + cot^{2} (θ)) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Join

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

\frac{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}{\frac{1}{s e c ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{sec ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec^{2} (θ)

sec^{2} (θ)

sec^{2} (θ)

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