Is csc(θ)=sec(θ)cot(θ) ?

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

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prove csc(θ)=sec(θ)cot(θ)

prove

csc (θ) = sec (θ) cot (θ)

True

Solution steps

csc (θ) = sec (θ) cot (θ)

Manipulating right side

sec (θ) cot (θ)

Express with sin, cos

cot (θ) sec (θ)

Use the basic trigonometric identity:

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

= \frac{cos ( θ )}{sin ( θ )} sec (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Cancel the common factor:

cos (θ)

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{csc ( x ) sin ( x )}{tan ( x )} = cot (x)

cos (- 2 x) = cosh^{2} (x) + sinh^{2} (x)

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

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

sec (2 x) + tan (2 x) = \frac{( cos ( x ) + sin ( x ) )}{( cos ( x ) - sin ( x ) )}