Is cot(θ)csc(θ)sec(θ)=csc^2(θ) ?

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

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prove cot(θ)csc(θ)sec(θ)=csc^2(θ)

prove

cot (θ) csc (θ) sec (θ) = csc^{2} (θ)

True

Solution steps

cot (θ) csc (θ) sec (θ) = csc^{2} (θ)

Manipulating left side

cot (θ) csc (θ) sec (θ)

Express with sin, cos

cot (θ) csc (θ) sec (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{1 \cdot 1}{sin ( θ ) sin ( θ )}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{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{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{csc ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) = \frac{2 tan ( \frac{x}{2} )}{1 - tan ^{2} ( \frac{x}{2} )}

4 sin (x) cos (x) cos (2 x) = sin (4 x)

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

- 2 tan^{2} (0) = 2 tan (0)

2 cot (x) sin (x) = 2 cos (x)

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