Is cot(θ)csc(θ)tan^2(θ)=sec(θ) ?

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

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prove cot(θ)csc(θ)tan^2(θ)=sec(θ)

Solution

prove

cot (θ) csc (θ) tan^{2} (θ) = sec (θ)

Solution

True

Solution steps

cot (θ) csc (θ) tan^{2} (θ) = sec (θ)

Manipulating left side

cot (θ) csc (θ) tan^{2} (θ)

Express with sin, cos

cot (θ) csc (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply the fraction rule:

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

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

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

sec (θ)

sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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